#

Multiple Choice Questions

One and More than One Correct

1. A magnetic field $B = B_0\hat{j}$ exists in the region $a < x < 2a$ and $B = -B_0\hat{j}$, in the region $2a < x < 3a$ where $B_0$ is a positive constant. A positive point charge moving with a velocity $\vec{v} = v_0\hat{i}$, where $v_0$ is a positive constant, enters the magnetic field at $x = a$. The trajectory of the charge in this region can be like, [JEE 2007 P-II]

Answer: (a)

Solution:

For $a < x < 2a$ path will be concave upward. For $2a < x < 3a$ path will be concave downward.

2. A particle of mass $m$ and charge $q$, moving with velocity $V$ enters Region II normal to the boundary as shown in the figure. Region II has a uniform magnetic field $B$ perpendicular to the plane of the paper. The length of the Region II is $\lambda$. Choose the correct choice(s). [JEE 2008 P-I]

a. The particle enters Region III only if its velocity $V > \frac{qB\lambda}{m}$

b. The particle enters Region III only if its velocity $V < \frac{qB\lambda}{m}$

c. Path length of the particle is Region II is maximum when velocity $V = \frac{qB\lambda}{m}$

d. Time spent in Region II is same for any velocity $V$ as long as the particle returns to Region I

Answer: (a, c, d)

3. The figure shows certain wire segments joined together to form a coplanar loop. The loop is placed in a perpendicular magnetic field in the direction going into the plane of the figure. The magnitude of the field increases with time. $I_1$ and $I_2$ are the currents in the segments $ab$ and $cd$. Then, [JEE 2009 P-I]

a. $I_1 > I_2$

b. $I_1 < I_2$

c. $I_1$ is in the direction $ba$ and $I_2$ is in the direction $cd$

d. $I_1$ is in the direction $ab$ and $I_2$ is in the direction $dc$

Answer: (d)

Solution:

According to Lenz’s law, current will be in anticlockwise sense as magnetic field is increasing into the plane of paper.

4. A thin flexible wire of length $L$ is connected to two adjacent fixed points and carries a current $I$ in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength $B$ going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is: [JEE 2010 P-I]

a. $IBL$

b. $\frac{IBL}{\pi}$

c. $\frac{IBL}{\pi^2}$

d. $\frac{IBL}{4\pi}$

Answer: (c)

Solution:

$$2T\sin\left(\frac{d\theta}{2}\right) = BiRd\theta \quad (\text{For } \theta \text{ small})$$ $$\Rightarrow Td\theta = BiRd\theta \Rightarrow T = BiR = \frac{IBL}{\pi^2}$$

5. Along insulated copper wire is closely wound as a spiral of '$N$' turns. The spiral has inner radius '$a$' and outer radius '$b$'. The spiral lies in the X-Y plane and a steady current '$I$' flows through the wire. The Z component of the magnetic field at the centre of the spiral is: [JEE 2011 P-II]

a. $\frac{\mu_0 NI}{2(b-a)} \ln\left(\frac{b}{a}\right)$

b. $\frac{\mu_0 NI}{2(b-a)} \ln\left(\frac{b+a}{b-a}\right)$

c. $\frac{\mu_0 NI}{2b} \ln\left(\frac{b}{a}\right)$

d. $\frac{\mu_0 NI}{2(b-a)} \ln\left(\frac{b+a}{b}\right)$

Answer: (a)

Solution:

$$B = \int_{a}^{b} \frac{\mu_0 I \left(\frac{N}{b-a} dr\right)}{2r} = \frac{\mu_0 NI}{2(b-a)} \ln\left(\frac{b}{a}\right)$$

6. An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement is/are true? [JEE 2011 P-I]

a. They will never come out of the magnetic field region

b. They will come out travelling along parallel paths

c. They will come out at the same time

d. They will come out at different times

Answer: (b, d)

Solution:

As shown in the figure they will come out after describing a semicircle along parallel paths. Also time taken $t = \frac{\pi m}{qB}$ which will be different for both.

7. A loop carrying current $I$ lies in the x-y plane as shown in the figure. The unit vector $\hat{k}$ is coming out of the plane of the paper. The magnetic moment of the current loop is: [JEE 2012 P-II]

a. $a^2 I \hat{k}$

b. $a^2 I \left(1 + \frac{\pi}{2}\right)\hat{k}$

c. $a^2 I \left(1 - \frac{\pi}{2}\right)\hat{k}$

d. $a^2 I (2\pi + 1)\hat{k}$

Answer: (b)

Solution:

$\vec{M} = I\vec{A}$

$$\vec{M} = I a^2 \left(1 + \frac{\pi}{2}\right)\hat{k}$$

8. An infinitely long hollow conducting cylinder with inner radius $R/2$ and outer radius $R$ carries a uniform current density along its length. The magnitude of the magnetic field, $\vec{B}$ as a function of the radial distance $r$ from the axis is best represented by. [JEE 2012 P-II]

Answer: (d)

Solution:

Let current be $I$. Current density $J = \frac{I}{\pi R^2 - \pi (R/2)^2} = \frac{4I}{3\pi R^2}$.

For $0 \le r \le \frac{R}{2}$, $B = 0$.

For $\frac{R}{2} < r \le R$, applying Ampere's Law:

$$B \cdot 2\pi r = \mu_0 \left[ \pi \left( r^2 - \frac{R^2}{4} \right) J \right]$$ $$B = \frac{\mu_0 I}{2\pi r} \left( \frac{4r^2 - R^2}{3R^2} \right)$$

At $r = \frac{R}{2}$, $B = 0$. As $r$ increases to $R$, $B = \frac{\mu_0 I}{2\pi R}$.

For $r \ge R$, $B = \frac{\mu_0 I}{2\pi r}$ ($\propto \frac{1}{r}$). This proves the continuity in the graph at $r = R$.

9. A particle of mass $M$ and positive charge $Q$, moving with a constant velocity $\vec{u}_1 = 4\hat{i}\text{ m/s}$, enters a region of uniform static magnetic field normal to the x-y plane. The region of the magnetic field extends from $x = 0$ to $x = L$ for all values of $y$. After passing through this region, the particle emerges on the other side after 10 milliseconds with a velocity $\vec{u}_2 = 2(\sqrt{3}\hat{i} + \hat{j})\text{ m/s}$. The correct statement(s) is (are): [JEE Adv. 2013 P-I]

a. The direction of the magnetic field is –z direction

b. The direction of the magnetic field is +z direction

c. The magnitude of the magnetic field $\frac{50M}{3Q}\pi$ units

d. The magnitude of the magnetic field is $\frac{100M}{3Q}\pi$ units

Answer: (a, c)

Solution:

Magnetic field is along –ve z-direction. Time taken in the magnetic field $t = \frac{\pi M}{6QB} = 10 \times 10^{-3}\text{ s}$, which gives $B = \frac{50\pi M}{3Q}$ units.

10. A steady current $I$ flow along an infinitely long hollow cylindrical conductor of radius $R$. This cylinder is placed coaxially inside an infinite solenoid of radius $2R$. The solenoid has $n$ turns per unit length and carries a steady current $I$. Consider a point $P$ at a distance $r$ from the common axis. The correct statement(s) is (are): [JEE Adv. 2013 P-II]

a. In the region $0 < r < R$, the magnetic field is non-zero

b. In the region $R < r < 2R$, the magnetic field is along the common axis

c. In the region $R < r < 2R$, the magnetic field is tangential to the circle of radius $r$, centered on the axis

d. In the region $r > 2R$, the magnetic field is non-zero

Answer: (a, d)

Solution:

Due to field of solenoid is non zero in region $0 < r < R$ and non zero in region $r > 2R$ due to conductor.

11. A circular loop of radius $3.0\text{ cm}$ lies parallel to a much bigger circular loop of radius $20\text{ cm}$. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is $15\text{ cm}$. If a current of $0.2\text{ A}$ flows through the smaller loop, then the flux linked with bigger loop is: [JEE Main 2013]

a. $1.91 \times 10^{-11}\text{ weber}$

b. $10^6 \times 10^{-11}\text{ weber}$

c. $3.3 \times 10^{-11}\text{ weber}$

d. $6.6 \times 10^{-9}\text{ weber}$

Answer: (a)

Solution:

Using the formula for mutual inductance and flux linked: $\Phi = M \cdot I \approx 1.91 \times 10^{-11}\text{ Wb}$.

12. Two short bar magnets of length $1\text{ cm}$ each have magnetic moments $1.20\text{ Am}^2$ and $1.00\text{ Am}^2$ respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by a distance of $20.0\text{ cm}$. The value of the resultant horizontal magnetic induction at the mid-point $O$ of the line joining their centre is close to (Horizontal component of earth’s magnetic induction is $3.6 \times 10^{-5}\text{ Wb m}^{-2}$) [JEE Main 2013]

a. $6.3 \times 10^{-5}\text{ Wb m}^{-2}$

b. $2.56 \times 10^{-4}\text{ Wb m}^{-2}$

c. $3.05 \times 10^{-4}\text{ Wb m}^{-2}$

d. $8.05 \times 10^{-4}\text{ Wb m}^{-2}$

Answer: (b)

Solution:

Resultant magnetic field $B_{\text{net}} = B_1 + B_2 + B_H = 2.56 \times 10^{-4}\text{ Wb m}^{-2}$.

13. The coercivity of a small magnet where the ferromagnet gets demagnetised is $3 \times 10^3\text{ Am}^{-1}$. The current required to be passed in a solenoid of length $10\text{ cm}$ and number of turns $100$, so that the magnet gets demagnetized when inside the solenoid, is: [JEE Main 2014]

a. $3\text{ A}$

b. $6\text{ A}$

c. $30\text{ mA}$

d. $60\text{ mA}$

Answer: (a)

Solution:

$$H = n i \Rightarrow i = \frac{H}{n} = \frac{3 \times 10^3}{\frac{100}{0.1}} = 3\text{ A}$$

14. A conductor lies along the z-axis at $-1.5 \le z < 1.5\text{ m}$ and carries a fixed current of $10.0\text{ A}$ in $-\hat{z}$ direction (see figure). For a field $\vec{B} = 3.0 \times 10^{-4} e^{-0.2x} \hat{x}\text{ T}$, find the power required to move the conductor at constant speed to $x = 2.0\text{ m}, y = 0\text{ m}$ in $5 \times 10^{-3}\text{ s}$. Assume parallel motion along the x-axis. [JEE Main 2014]

a. $14.85\text{ W}$

b. $29.7\text{ W}$

c. $1.57\text{ W}$

d. $2.97\text{ W}$

Answer: (d)

Solution:

Power $P = \frac{W}{T} = \frac{\int F dx}{T} = 2.97\text{ W}$.

15. A conductor (shown in the figure) carrying constant current $I$ is kept in the x-y plane in a uniform magnetic field $\vec{B}$. If $F$ is the magnitude of the total magnetic force acting on the conductor, then the correct statement(s) is(are): [JEE Adv. 2015 P-I]

a. If $\vec{B}$ is along $\hat{z}$, $F \propto (L + R)$

b. If $\vec{B}$ is along $\hat{x}$, $F = 0$

c. If $\vec{B}$ is along $\hat{y}$, $F \propto (L + R)$

d. If $\vec{B}$ is along $\hat{z}$, $F = 0$

Answer: (a, b, c)

16. A rectangular loop of sides $10\text{ cm}$ and $5\text{ cm}$ carrying a current $I$ of $12\text{ A}$ is placed in different orientations as shown in the figures below: If there is a uniform magnetic field of $0.3\text{ T}$ in the positive z direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium? [JEE Main 2015]

a. a. and b., respectively

b. a. and c., respectively

c. b. and d., respectively

d. b. and c., respectively

Answer: (c)

Solution:

In both option (b) and (d), $\vec{F}_{\text{net}} = 0, \vec{\tau}_{\text{net}} = 0$.

17. Two identical wires A and B, each of length $l$, carry the same current $I$. Wire A is bent into a circle of radius $R$ and wire B is bent to form a square of side '$a$'. If $B_A$ and $B_B$ are the values of magnetic field at the centres of the circle and square respectively, then the ratio $B_A / B_B$ is: [JEE Main 2016]

a. $\frac{\pi^2}{8}$

b. $\frac{\pi^2}{16\sqrt{2}}$

c. $\frac{\pi^2}{16}$

d. $\frac{\pi^2}{8\sqrt{2}}$

Answer: (d)

Solution:

Magnetic field at centre of circle $B_A = \frac{\mu_0 I}{2R} = \frac{\mu_0 \pi I}{l}$. Magnetic field at centre of square $B_B = \frac{4\sqrt{2}\mu_0 I}{\pi l}$. Ratio $\frac{B_A}{B_B} = \frac{\pi^2}{8\sqrt{2}}$.

18. Hysteresis loops for two magnetic materials A and B are given below: These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use: [JEE Main 2016]

a. A for electric generators and transformers.

b. A for electromagnets and B for electric generators.

c. A for transformers and B for electric generators.

d. B for electromagnets and transformers.

Answer: (d)

Solution:

Since area of Hysteresis curve of (b) is smaller it is suitable for electromagnet and transformer.

19. A conducting loop in the shape of a right angled isosceles triangle of height $10\text{ cm}$ is kept such that the $90^\circ$ vertex is very close to an infinitely long conducting wire (see the figure). The wire is electrically insulated from the loop. The hypotenuse of the triangle is parallel to the wire. The current in the triangular loop is in counterclockwise direction and increased at a constant rate of $10\text{ A s}^{-1}$. Which of the following statement(s) is (are) true? [JEE Adv. 2016 P-I]

a. The magnitude of induced emf in the wire is $\frac{\mu_0}{\pi}$ volt

b. If the loop is rotated at a constant angular speed about the wire, an additional emf of $\frac{\mu_0}{\pi}$ volt is induced in the wire

c. The induced current in the wire is in opposite direction to the current along the hypotenuse

d. There is a repulsive force between the wire and the loop

Answer: (a, d)

Solution:

Coefficient of mutual inductance $M = \frac{\mu_0 \lambda}{\pi}$. $\text{emf} = M \frac{dI}{dt} = \frac{\mu_0}{\pi}$ volts. The loop should move away according to Lenz's law. So, there must be repulsive force.

20. A rigid wire loop of square shape having side of length $L$ and resistance $R$ is moving along the x-axis with a constant velocity $v_0$ in the plane of the paper. At $t = 0$, the right edge of the loop enters a region of length $3L$ where there is a uniform magnetic field $B_0$ into the plane of the paper; as shown in the figure. For sufficiently large $v_0$, the loop eventually crosses the region. Let $x$ be the location of the right edge of the loop. Let $v(x)$, $I(x)$ and $F(x)$ represent the velocity of the loop, current in the loop, and force on the loop, respectively, as a function of $x$. Counter-clockwise current is taken as positive. [JEE Adv. 2016 P-II]

Which of the following schematic plot(s) is (are) correct? (Ignore gravity)

Answer: (c, d)

21. In a coil of resistance $100\ \Omega$, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is [JEE Main 2017]

a. $225\text{ Wb}$

b. $250\text{ Wb}$

c. $275\text{ Wb}$

d. $220\text{ Wb}$

Answer: (b)

Solution:

$\Delta\phi = qR = (\text{Area of current-time graph}) \times R = 2.5 \times 100 = 250\text{ Wb}$.

22. A magnetic needle of magnetic moment $6.7 \times 10^{-2}\text{ Am}^2$ and moment of inertia $7.5 \times 10^{-6}\text{ kg m}^2$ is performing simple harmonic oscillations in a magnetic field of $0.01\text{ T}$. Time taken for 10 complete oscillations is: [JEE Main 2017]

a. $8.89\text{ s}$

b. $6.98\text{ s}$

c. $8.76\text{ s}$

d. $6.65\text{ s}$

Answer: (d)

Solution:

$T = 2\pi \sqrt{\frac{I}{MB}} = 0.665\text{ s}$. For 10 oscillations, time $= 10 T = 6.65\text{ seconds}$.

23. A circular insulated copper wire loop is twisted to form two loops of area $A$ and $2A$ as shown in the figure. At the point crossing the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field $\vec{B}$ points into the plane of the paper. At $t = 0$, the loop starts rotating about the common diameter as axis with a constant angular velocity $\omega$ in the magnetic field. Which of the following options is/are correct? [JEE Adv. 2017 P-I]

a. The emf induced in the loop is proportional to the sum of the areas of the two loops.

b. The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper

c. The net emf induced due to both the loops is proportional to $\cos\omega t$

d. The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude of maximum emf induced in the smaller loop alone.

Answer: (b, d)

Solution:

Rate of change of flux is maximum when $\sin\omega t = \pm 1$, i.e., plane of loop is perpendicular to the plane of paper.

24. A symmetric star shaped conducting wire loop is carrying a steady state current $I$ as shown in the figure. The distance between the diametrically opposite vertices of the star is $4a$. The magnitude of the magnetic field at the center of the loop is [JEE Adv. 2017 P-II]

a. $\frac{\mu_0 I}{4\pi a} [6\sqrt{3} - 1]$

b. $\frac{\mu_0 I}{4\pi a} [6\sqrt{3} + 1]$

c. $\frac{\mu_0 I}{4\pi a} [3\sqrt{3} - 1]$

d. $\frac{\mu_0 I}{4\pi a} [3\sqrt{2} - 3]$

Answer: (a)

Solution:

Magnetic field at the center of the loop due to all 12 identical sections is additive in nature: $B_{\text{net}} = \frac{\mu_0 I}{4\pi a} [6\sqrt{3} - 1]$.

25. A uniform magnetic field $B$ exists in the between $x = 0$ and $x = \frac{3}{2}R$ (region 2 in the figure) pointing normally into the plane of the paper. A particle with charge $+Q$ and momentum $p$ directed along x-axis enters region 2 from region 1 at point $P_1(0, -R/2)$. Which of the following option(s) is/are correct? [JEE Adv. 2017 P-II]

a. When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the change in its momentum between point $P_1$ and the farthest point from y-axis is $\sqrt{3}p$

b. For $p = \frac{8}{13}QBR$, the particle will enter region 3 through the point $P_2$ on x-axis

c. For $p > \frac{2}{3}QBR$, the particle will re-enter region 1

d. For a fixed $B$, particles of same charge $Q$ and same velocity $v$, the distance between the point $P_1$ and the point of re-entry into region 1 is inversely proportional to the mass of the particle

Answer: (b, c)

26. The dipole moment of a circular loop carrying a current $I$, is $m$ and the magnetic field at the centre of the loop is $B_1$. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is $B_2$. The ratio $\frac{B_1}{B_2}$ is [JEE Main 2018]

a. $\frac{1}{2}$

b. $2$

c. $3$

d. $\sqrt{2}$

Answer: (d)

Solution:

$m = I A = I \pi R^2$. When $m$ is doubled at constant $I$, $R' = \sqrt{2} R$. $B_1 = \frac{\mu_0 I}{2R}$, $B_2 = \frac{\mu_0 I}{2R'} = \frac{B_1}{\sqrt{2}}$, so $\frac{B_1}{B_2} = \sqrt{2}$.

27. Two infinitely long straight wires lie in the xy-plane along the lines $x = \pm R$. The wire located at $x = +R$ carries a constant current $I_1$ and the wire located at $x = -R$ carries a constant current $I_2$. A circular loop of radius $R$ is suspended with its centre at $(0,0, \sqrt{3}R)$ and in a plane parallel to the xy-plane. This loop carries a constant current $I$ in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the $+\hat{j}$ direction. Which of the following statements regarding the magnetic field $\vec{B}$ is (are) true? [JEE Adv. 2018 P-I]

a. If $I_1 = I_2$, then $\vec{B}$ cannot be equal to zero at the origin $(0,0,0)$

b. If $I_1 > 0$ and $I_2 < 0$, then $\vec{B}$ can be equal to zero at the origin $(0, 0, 0)$

c. If $I_1 < 0$ and $I_2 > 0$, then $\vec{B}$ can be equal to zero at the origin $(0, 0, 0)$

d. If $I_1 = I_2$, then the z-component of the magnetic field at the centre of the loop is $-\left(\frac{\mu_0 I}{2R}\right)$

Answer: (a, b, d)

28. Two very long, straight, and insulated wires are kept at $90^\circ$ angle from each other in xy-plane as shown in the figure. These wires carry currents of equal magnitude $I$, whose directions are shown in the figure. The net magnetic field at point $P$ will be: [JEE Main 2019]

a. zero

b. $\frac{\mu_0 I}{\pi d} \hat{z}$

c. $\frac{\mu_0 I}{2\pi d} (\hat{x} - \hat{y})$

d. $\frac{\mu_0 I}{2\pi d} (\hat{x} + \hat{y})$

Answer: (a)

Solution:

Magnetic field at point P: $\vec{B}_{\text{net}} = \frac{\mu_0 I}{2\pi d}(-\hat{k} + \hat{k}) = 0$.

29. Two magnetic dipoles X and Y are placed at a separation $d$, with their axes perpendicular to each other. The dipole moment of Y is twice that of X. A particle of charge $q$ is passing, through their midpoint $P$, at angle $\theta = 45^\circ$ with the horizontal line, as shown in figure. What would be the magnitude of force on the particle at that instant? ($d$ is much larger than the dimensions of the dipole) [JEE Main 2019]

a. $\frac{\mu_0 M q v}{\pi (d/2)^3 3\sqrt{2}}$

b. $\frac{\mu_0 M q v}{\pi (d/2)^3 2}$

c. $\frac{\mu_0 M q v}{4\pi (d/2)^3}$

d. $0$

Answer: (d)

Solution:

Velocity of charge and $\vec{B}_{\text{net}}$ are parallel, so by $\vec{F} = q(\vec{v} \times \vec{B})$, force on charge particle is zero.

30. A circular loop of radius $r$ having $N$ number of turns carries current $I$ is placed in a uniform magnetic field $\vec{B}$ parallel to the plane of the loop. The torque on the loop is: [JEE Main 2019]

a. $\pi NI r B$

b. $2\pi^2 NI r^2 B$

c. $\pi^2 NI r^2 B$

d. $\pi NI r^2 B$

Answer: (d)

Solution:

$\vec{\tau} = \vec{M} \times \vec{B} \Rightarrow \tau = MB \sin 90^\circ = (N I \pi r^2) B = \pi NI r^2 B$.

31. A conducting slider of resistance $R(10\ \Omega)$, mass $50\text{ g}$ and length $10\text{ cm}$ is kept on a U-shaped frame as shown in figure. There is uniform magnetic field ($B = 0.1\text{ T}$) perpendicular to plane of frame. The slider is attached to a spring ($K = 0.5\text{ N/m}$). It is displace by $A_0$ and released time is which amplitude become $A_0/e$ is [JEE Main 2019]

a. $9000\text{ s}$

b. $10000\text{ s}$

c. $12000\text{ s}$

d. $15000\text{ s}$

Answer: (b)

Solution:

$t = \frac{2mR}{B^2 l^2} = \frac{2 \times 50 \times 10^{-3} \times 10}{(0.1)^2 \times (0.1)^2} = 10000\text{ s}$.

32. One of the two identical conducting wires of length $L$ is bent in the form of a circular loop and the other one into a circular coil of $N$ identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop ($B_L$) to that at the centre of the coil ($B_C$) i.e. $\frac{B_L}{B_C}$ will be: [JEE Main 2019]

a. $\frac{1}{N}$

b. $N^2$

c. $\frac{1}{N^2}$

d. $N$

Answer: (c)

Solution:

$B_L = \frac{\mu_0 i}{2R}$, $B_C = \frac{\mu_0 N i}{2r}$. Since $2\pi R = 2\pi N r \Rightarrow R = N r$. Thus, $\frac{B_L}{B_C} = \frac{r}{N R} = \frac{1}{N^2}$.

33. A particle having the same charge as of electron moves in a circular path of radius $0.5\text{ cm}$ under the influence of a magnetic field of $0.5\text{ T}$. if an electric field of $100\text{ V/m}$ makes it to move in a straight path, then the mass of the particle is: (Given charge of electron = $1.6 \times 10^{-19}\text{C}$) [JEE Main 2019]

a. $2.0 \times 10^{-24}\text{ kg}$

b. $1.6 \times 10^{-19}\text{ kg}$

c. $1.6 \times 10^{-27}\text{ kg}$

d. $9.1 \times 10^{-31}\text{ kg}$

Answer: (a)

Solution:

$m = \frac{q B^2 R^2}{E} = 2.0 \times 10^{-24}\text{ kg}$.

34. A current loop, having two circular arcs joined by two radial lines is shown in the figure. It carries a current of $10\text{A}$. The magnetic field at point $O$ will be close: [JEE Main 2019]

a. $1.0 \times 10^{-5}\text{ T}$

b. $1.0 \times 10^{-5}\text{ T}$

c. $1.0 \times 10^{-7}\text{ T}$

d. $1.0 \times 10^{-8}\text{ T}$

Answer: (a)

Solution:

Using the formula for magnetic field due to circular arcs at the center, $B = \frac{\mu_0 I \theta}{4\pi r}$, $B \approx 1.0 \times 10^{-5}\text{ T}$.

35. A bar magnet is demagnetised by inserting it inside a solenoid of length $0.2\text{ m}$, $100$ turns, and carrying a current of $5.2\text{A}$. The coercivity of the bar magnet is: [JEE Main 2019]

a. $1200\text{ A/m}$

b. $2600\text{ A/m}$

c. $520\text{ A/m}$

d. $285\text{ A/m}$

Answer: (b)

Solution:

$H = \frac{N i}{l} = \frac{100 \times 5.2}{0.2} = 2600\text{ A/m}$.

36. An infinitely long current carrying wire and a small current carrying loop are in the plane of the paper as shown. The radius of the loop is $a$ and distance of its centre from the wire is $d$ ($d \gg a$). If the loop applies a force $F$ on the wire then: [JEE Main 2019]

a. $F \propto \left(\frac{a}{d}\right)^3$

b. $F \propto \frac{a}{d}$

c. $F \propto \left(\frac{a}{d}\right)^2$

d. $F = 0$

Answer: (a)

Solution:

$\infty$ long wire

Equivalent dipole of given loop

$$F = m \frac{dB}{dr}$$ $$\text{New } B = \frac{\mu_0}{2\pi} \frac{dI}{dx}$$ $$\frac{dB}{dx} \propto \frac{1}{x^3}$$ $$\text{So } F \propto \frac{M}{x^3} \quad [\because M = \text{NIA}]$$ $$\therefore F \propto \left(\frac{a}{d}\right)^3$$

37. At some location on earth the horizontal component of earth's magnetic field is $18 \times 10^{-6}\text{ T}$. At this location, magnetic needle of length $0.12\text{ m}$ and pole strength $1.8\text{ Am}$ is suspended from its mid-point using a thread, it makes $45^\circ$ angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is: [JEE Main 2019]

a. $3.6 \times 10^{-5}\text{ N}$

b. $6.5 \times 10^{-5}\text{ N}$

c. $1.3 \times 10^{-5}\text{ N}$

d. $1.8 \times 10^{-5}\text{ N}$

Answer: (b)

Solution:

$$\mu B l \sin 45^\circ = F \frac{l}{2} \sin 45^\circ$$ $$F = 2\mu B$$

38. A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are $T_h$ and $T_c$ respectively, then: [JEE Main 2019]

a. $T_h = 0.5 T_c$

b. $T_h = 2 T_c

c. $T_h = 1.5 T_c$

d. $T_h = T_c$

Answer: (d)

Solution:

$$T = 2\pi \sqrt{\frac{I}{MB}}$$ $$T_h = 2\pi \sqrt{\frac{mR^2}{2MB}} = 2\pi \sqrt{\frac{mR^2}{2(2M')B}}$$ $$T_c = 2\pi \sqrt{\frac{\frac{1}{2}mR^2}{M'B}} = 2\pi \sqrt{\frac{mR^2}{2M'B}}$$ $$T_h = T_c$$

39. A solid metal cube of edge length $2\text{ cm}$ is moving in a positive y direction at a constant speed of $6\text{ m/s}$. There is a uniform magnetic field of $0.1\text{ T}$ in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis is: [JEE Main 2019]

a. $6\text{ mV}$

b. $1\text{ mV}$

c. $12\text{ mV}$

d. $2\text{ mV}$

Answer: (c)

Solution:

Potential difference between two faces perpendicular to x-axis will be:

$$\Delta V = (\vec{v} \times \vec{B}) \cdot \vec{l} = v B l = (6)(0.1)(0.02) = 0.012\text{ V} = 12\text{ mV}$$

40. A magnet of total magnetic moment $2\hat{i}\text{ Am}^2$ is placed in a time varying magnetic field, $B\hat{i} (\cos\omega t)$ where $B = 1\text{ Tesla}$ and $\omega = 0.125\text{ rad/s}$. The work done for reversing the direction of the magnetic moment at $t = 1\text{ second}$ is: [JEE Main 2019]

a. $0.007\text{ J}$

b. $0.014\text{ J}$

c. $0.01\text{ J}$

d. $0.028\text{ J}$

Answer: (b)

Solution:

$$\text{Work done, } W = -\Delta U = -\vec{\mu} \cdot \vec{B}_{\text{final}} + \vec{\mu} \cdot \vec{B}_{\text{initial}} = 2\mu B [1 - \cos(\omega t)]$$ $$W = 2 \times 2 \times 1 \times (1 - \cos(0.125 \times 1)) = 4(1 - \cos 0.125) \approx 4(1 - 0.992) = 0.032 \dots$$

41. A paramagnetic substance in the form of a cube with sides $1\text{ cm}$ has a magnetic dipole moment of $20 \times 10^{-6}\text{ J/T}$ when a magnetic intensity of $60 \times 10^3\text{ A/m}$ is applied. Its magnetic susceptibility is: [JEE Main 2019]

a. $2.3 \times 10^{-2}$

b. $3.3 \times 10^{-2}$

c. $3.3 \times 10^{-4}$

d. $4.3 \times 10^{-2}$

Answer: (c)

Solution:

$$\chi = \frac{I}{H}$$ $$\text{Magnetic moment } I = \frac{\text{Magnetic Moment}}{\text{Volume}} = \frac{20 \times 10^{-6}}{(10^{-2})^3} = \frac{20 \times 10^{-6}}{10^{-6}} = 20\text{ A/m}$$ $$\chi = \frac{20}{60 \times 10^3} = 0.33 \times 10^{-3} = 3.3 \times 10^{-4}$$

42. A galvanometer having a resistance of $20\ \Omega$ and $30$ divisions on both sides has figure of merit $0.005\text{ ampere/division}$. The resistance that should be connected in series such that it can be used as a voltmeter up to $15\text{ volt}$, is: [JEE Main 2019]

a. $80\ \Omega$

b. $120\ \Omega$

c. $125\ \Omega$

d. $100\ \Omega$

Answer: (a)

Solution:

$$R_g = 20\ \Omega$$ $$I_g = (\text{Figure of merit}) \times (\text{Total divisions}) = 0.005 \times 30 = 0.15\text{ A}$$ $$V = I_g(R_g + R)$$ $$15 = 0.15(20 + R)$$ $$100 = 20 + R \implies R = 80\ \Omega$$

43. In an experiment electrons are accelerated, from rest, by applying a voltage of $500\text{ V}$. Calculate the radius of the path if a magnetic field $100\text{ mT}$ is then applied. [Charge of the electron $= 1.6 \times 10^{-19}\text{ C}$, Mass of the electron $= 9.1 \times 10^{-31}\text{ kg}$] [JEE Main 2019]

a. $7.5 \times 10^{-4}\text{ m}$

b. $7.5 \times 10^{-3}\text{ m}$

c. $7.5\text{ m}$

d. $7.5 \times 10^{-2}\text{ m}$

Answer: (a)

Solution:

$$r = \frac{\sqrt{2m q V}}{q B} = \frac{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 500}}{1.6 \times 10^{-19} \times 100 \times 10^{-3}} \approx 7.5 \times 10^{-4}\text{ m}$$

44. A paramagnetic material has $10^{28}\text{ atoms/m}^3$. Its magnetic susceptibility at temperature $350\text{ K}$ is $2.8 \times 10^{-4}$. Its susceptibility at $300\text{ K}$ is: [JEE Main 2019]

a. $3.672 \times 10^{-4}$

b. $3.726 \times 10^{-4}$

c. $3.267 \times 10^{-4}$

d. $2.672 \times 10^{-4}$

Answer: (c)

Solution:

Curie's law for paramagnetic substance: $\chi \propto \frac{1}{T}$

$$\frac{\chi_1}{\chi_2} = \frac{T_2}{T_1}$$ $$\frac{2.8 \times 10^{-4}}{\chi_2} = \frac{300}{350}$$ $$\chi_2 = \frac{2.8 \times 10^{-4} \times 350}{300} = 3.267 \times 10^{-4}$$

45. A $10\text{ m}$ long horizontal wire extends from North East to South West. It is falling with a speed of $5.0\text{ ms}^{-1}$, at right angles to the horizontal component of the earth's magnetic field, of $0.3 \times 10^{-4}\text{ Wb/m}^2$. The value of the induced emf in wire is: [JEE Main 2019]

a. $2.5 \times 10^{-3}\text{ V}$

b. $1.1 \times 10^{-3}\text{ V}$

c. $0.3 \times 10^{-3}\text{ V}$

d. $1.5 \times 10^{-3}\text{ V}$

Answer: (d)

Solution:

$$\text{Induced emf} = Bv l = 0.3 \times 10^{-4} \times 5 \times 10 = 1.5 \times 10^{-3}\text{ V}$$

46. A galvanometer, whose resistance is $50\ \Omega$, has $25$ divisions in it. When a current of $4 \times 10^{-4}\text{ A}$ passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range $2.5\text{ V}$, it should be connected to a resistance of: [JEE Main 2019]

a. $6250\ \Omega$

b. $250\ \Omega$

c. $200\ \Omega$

d. $6200\ \Omega$

Answer: (c)

Solution:

$$I_g = 25 \times 4 \times 10^{-4} = 10^{-2}\text{ A}$$ $$V = I_g(R_g + R)$$ $$2.5 = 10^{-2}(50 + R)$$ $$250 = 50 + R \implies R = 200\ \Omega$$

47. The galvanometer deflection, when key $K_1$ is closed but $K_2$ is open, equals $\theta_0$ (see figure). On closing $K_2$ also and adjusting $R_2$ to $5\ \Omega$, the deflection in galvanometer becomes $0.5\theta_0$. The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]: [JEE Main 2019]

a. $12\ \Omega$

b. $25\ \Omega$

c. $5\ \Omega$

d. $22\ \Omega$

Answer: (d)

Solution:

Case (i): $\theta_0 = \frac{E}{R_1 + R_g} C = \frac{E}{220 + R_g} C$

Case (ii): $0.5\theta_0 = \frac{E}{R_1 + \frac{R_g R_2}{R_g + R_2}} \cdot \frac{R_2}{R_g + R_2} C = \frac{E}{220 + \frac{5 R_g}{5 + R_g}} \cdot \frac{5}{5 + R_g} C$

Solving yields $R_g = 22\ \Omega$.

48. A proton and an $\alpha$-particle (with their masses in the ratio of $1:4$ and charges in the ratio of $1:2$) are accelerated from rest through a potential difference $V$. If a uniform magnetic field ($B$) is set up perpendicular to their velocities, the ratio of the radii $r_p : r_\alpha$ of the circular paths described by them will be: [JEE Main 2019]

a. $1: \sqrt{2}$

b. $1: 2$

c. $1 : \sqrt{3}$

d. $1: 3$

Answer: (a)

Solution:

$$r = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$$ $$\frac{r_p}{r_\alpha} = \sqrt{\frac{m_p}{m_\alpha} \cdot \frac{q_\alpha}{q_p}} = \sqrt{\frac{1}{4} \times \frac{2}{1}} = \frac{1}{\sqrt{2}}$$

49. As shown in the figure, two infinitely long, identical wires are bent by $90^\circ$ and placed in such a way that the segments $LP$ and $QM$ are along the x-axis, while segments $PS$ and $QN$ are parallel to the y-axis. If $OP = OQ = 4\text{ cm}$, and the magnitude of the magnetic field at $O$ is $10^{-4}\text{ T}$, and the two wires carry equal currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at $O$ will be ($\mu_0 = 4\pi \times 10^{-7}\text{ NA}^{-2}$): [JEE Main 2019]

a. $40\text{ A}$, perpendicular into the page

b. $40\text{ A}$, perpendicular out of the page

c. $20\text{ A}$, perpendicular out of the page

d. $20\text{ A}$, perpendicular into the page

Answer: (d)

Solution:

Magnetic field at $O$ is due to $PS$ and $QN$ only (since semi-infinite segments along x-axis contribute zero field at $O$).

$$B_0 = B_{PS} + B_{QN} = 2 \times \frac{\mu_0 i}{4\pi d} = \frac{\mu_0 i}{2\pi d}$$ $$10^{-4} = \frac{4\pi \times 10^{-7} \times i}{2\pi \times 0.04}$$ $$i = 20\text{ A}$$

Direction by right-hand rule is perpendicular into the page.

50. A galvanometer of resistance $10\ \Omega$ and maximum current of $2\ \mu\text{A}$ is converted into voltmeter of range $100\text{ mV}$ and when converted into ammeter then range is $1\text{ mA}$. When these voltmeter and ammeter are connected by a (ideal) battery in series with a resistance of $R = 1000\ \Omega$, then? [JEE Adv. 2019 P-I]

a. Measured value of $R$ is between $978\ \Omega$ and $996\ \Omega$

b. Resistance of voltmeter $10^5\ \Omega$

c. Shunt resistance is $20\text{ m}\Omega$

d. If the ideal battery is replaced by non ideal battery with internal resistance of $5\ \Omega$ then $R$ will be $> 1000\ \Omega$

Answer: (a, c)

Solution:

Voltmeter range $V = I_g(R_g + R_v) \implies 100 \times 10^{-3} = 2 \times 10^{-6}(10 + R_v) \implies R_v \approx 50000\ \Omega = 5 \times 10^4\ \Omega$.

Shunt resistance $S = \frac{I_g R_g}{I - I_g} = \frac{2 \times 10^{-6} \times 10}{10^{-3} - 2 \times 10^{-6}} \approx 20\text{ m}\Omega$.

51. Conducting wire of parabolic shape, initially $y = x^2$ is moving with velocity $\vec{v} = v_0\hat{i}$ is a non-uniform magnetic field $\vec{B} = B_0 \left(1 + \frac{y}{L}\right)\hat{k}$ as shown in figure. If $v_0, B_0, L$ \& $B$ are +ve constants and $\Delta\phi$ is potential difference develop between the ends of wire, then correct statement(s) is/are: [JEE Adv. 2019 P-I]

a. $\Delta\phi = B_0 V_0 L$ for $\beta = 0$

b. $\Delta\phi = \frac{4}{3} B_0 V_0 L$ for $\beta = 2$

c. $\Delta\phi$ is proportional to the length of wire projected on y-axis

d. $\Delta\phi$ remains same if the parabolic wire is replaced by a straight wire, $y = x$, initially of length $2\ell$

Answer: (b, c, d)

Solution:

Motional emf $\varepsilon = \int (\vec{v} \times \vec{B}) \cdot d\vec{l} = \int v_0 B_0 \left(1 + \frac{y}{L}\right) dy$. Evaluating for the given limits yields the correct options.

Matrix Match/Column Match

52. Some physical quantities are given in Column I and some possible SI units in which these quantities may be expressed are given in Column II. Match the physical quantities in Column I with the units in Column II. [JEE 2007 P-I]

Column I:

(A) $\frac{GM_e}{M_s}$

(B) $\frac{RT}{M}$

(D) $\frac{GMe}{R_e}$

Column II:

1. $(\text{volt})(\text{coulomb})(\text{metre})$

2. $(\text{kilogram})(\text{metre})^3(\text{second})^{-2}$

3. $(\text{metre})^2(\text{second})^{-2}$

4. $(\text{farad})(\text{volt})^2(\text{kg})^{-1}$

Answer: (a)

Solution:

A$\rightarrow$1, 2; B$\rightarrow$3, 4; C$\rightarrow$3, 4; D$\rightarrow$3, 4

53. Column I give certain situations in which a straight metallic wire of resistance $R$ is used and Column II gives some resulting effects. Match the statements in Column I with the statements in Column II. [JEE 2007 P-I]

Answer: (a)

Solution:

A$\rightarrow$2 ; B$\rightarrow$3, 4; C$\rightarrow$4; D$\rightarrow$1, 2, 3

54. Two wires each carrying a steady current $I$ are shown in four configurations in Column I. Some of the resulting effects are described in Column II. Match the statements in Column I with the statements in Column II. [JEE 2007 P-II]

Answer: (a)

Solution:

A$\rightarrow$2, 3; B$\rightarrow$1; C$\rightarrow$2, 3; D$\rightarrow$2

Paragraphs

Paragraph for Question Nos. 55 to 57

55. In which case would the particle move in a straight line along the negative direction of y-axis (i.e. move along $-\hat{y}$)? [JEE Adv. 2017 P-I]

Answer: (c)

Solution:

To move in a straight line along the negative direction of y-axis, both $\vec{E}$ and $\vec{B}$ must be along y axis and initial velocity must be either zero or along -y axis. Hence, option (c) is the only correct option.

56. In which case will the particle move in a straight line with constant velocity? [JEE Adv. 2017 P-I]

Answer: (a)

Solution:

To move with constant velocity, net force must be zero i.e., $q(\vec{v} \times \vec{B} + \vec{E}) = 0$. Such combination is only possible corresponding to option (a).

57. In which case will the particle describe a helical path with axis along the positive z direction? [JEE Adv. 2017 P-I]

Answer: (c)

Solution:

To describe a helical path with axis along $+z$ direction, electric field must be along z direction and $\vec{v}$ and $\vec{B}$ must be normal to z. This is possible only for (c) in given choices.

Paragraph for Question Nos. 58 to 59

58. The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change, is: [JEE Adv. 2013 P-II]

a. $\frac{1}{4}BR$

b. $\frac{1}{2}BR$

c. $BR$

d. $2BR$

Answer: (b)

Solution:

$$E(2\pi R) = \pi R^2 \frac{dB}{dt} \implies E = \frac{R}{2} \frac{dB}{dt}$$

59. The change in the magnetic dipole moment associated with the orbit, at the end of time interval of the magnetic field change, is: [JEE Adv. 2013 P-II]

a. $\gamma BQR + 2$

b. $\frac{1}{2} BQR - \gamma$

c. $\gamma \frac{BQR}{2}$

d. $2\gamma BQR$

Answer: (c)

Solution:

$$\Delta\mu = \gamma \Delta L = \gamma \left(\tau \Delta t\right) = \gamma \left(\frac{1}{2}QR^2 B\right)$$

Paragraph for Question Nos. 60 to 61

60. When wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height above the loop. In that case [JEE Adv. 2014 P-II]

a. current in wire 1 and 2 is the direction PQ and RS, respectively and $h \approx 1.2a$

b. current in wire 1 and wire 2 is the direction PQ and SR, respectively and $h \approx 1.2a$

c. current in wire 1 and wire 2 is the direction PQ and SR, respectively and $h \approx 1.2a$

d. current in wire 1 and wire 2 is the direction PQ and RS, respectively and $h \approx 1.2a$

Answer: (c)

Solution:

Net magnetic field is zero when fields due to wires and loop cancel each other out.

61. Consider and the loop is rotated about its diameter parallel to the wires by $30^\circ$ from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be: [JEE Adv. 2014 P-II]

Answer: (b)

Solution:

$$\tau = MB \sin\theta$$

Integer and Subjective

62. In a moving coil galvanometer, torque on the coil can be expressed as $\tau = ki$, where $i$ is current through the wire and $k$ is constant. The rectangular coil of the galvanometer having numbers of turns $N$, area $A$ and moment of inertia $I$ is placed in magnetic field $B$. Find: [JEE 2005]

a. $k$ in terms of given parameters $N, I, A$ and $B$.

b. the torsional constant of the spring, if a current $i_0$ produces a deflection of $\pi/2$ in the coil.

c. the maximum angle through which coil is deflected, if charge $Q$ is passed through the coil almost instantaneously.

Solution:

a. $k = NAB$

b. $C = \frac{2NABi_0}{\pi}$

c. $\theta_{\max} = \frac{NABQ}{\sqrt{IC}}$

63. A steady current $I$ goes through a wire loop $PQR$ having shape of a right angle triangle with $PQ = 3$, $PR = 4$ and $QR = 5$. If the magnitude of the magnetic field at $P$ due to this loop is $\frac{\mu_0 I}{4\pi} \frac{k}{x}$, find the value of $k$. [JEE 2009 P-II]

Answer: 7

Solution:

$$B = \frac{\mu_0 I}{4\pi x} (\cos 53^\circ + \cos 37^\circ)$$

64. A long circular tube of length $10\text{ m}$ radius $0.3\text{ m}$ carries a current $I$ along its curved surface as shown. A wire-loop of resistance $0.005\text{ ohm}$ and of radius $0.1\text{ m}$ is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as $I = I_0 \cos(300t)$ where $I_0$ is constant. If the magnetic moment of the loop is $N I_0 t \mu_0 \sin(300t)$, then 'N' is: [JEE 2011 P-I]

Answer: 6

Solution:

Using Faraday's law and magnetic flux linkage through the loop, $N = 6$.

65. A cylindrical cavity of diameter $a$ exists inside a cylinder of diameter $2a$ as shown in the figure. Both the cylinder and the cavity are infinitely long. A uniform current density $J$ flows along the length. If the magnitude of the magnetic field at the point $P$ is given by $\frac{N}{12} \mu_0 a J$, then the value of $N$ is. [JEE 2012 P-I]

Answer: 5

Solution:

Using superposition principle for solid cylinder minus cavity cylinder, $N = 5$.

66. A circular wire loop of a radius $R$ is placed in the x–y plane centered at the origin $O$. A square loop of side $a$ ($a \ll R$) having two turns is placed with its centre at $z = \sqrt{3} R$ along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of $45^\circ$ with respect to the z–axis. If the mutual inductance between the loops is given by $\frac{\mu_0 a^2}{p R}$, then the value of $p$ is. [JEE 2012 P-I]

Answer: 7

Solution:

Calculate magnetic flux linked through the square loop due to current in the circular loop to find mutual inductance, giving $p = 7$.

67. A uniform circular disc of mass $50\text{ kg}$ and radius $0.4\text{ m}$ is rotating with an angular velocity of $10\text{ rad s}^{-1}$ about its own axis, which is vertical. Two uniform circular rings, each of mass $6.25\text{ kg}$ and radius $0.2\text{ m}$, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in $\text{rad s}^{-1}$) of the system is: [JEE Adv. 2013 P-I]

Answer: 8

Solution:

Using conservation of angular momentum about the vertical axis: $I_1 \omega_1 = I_2 \omega_2 \implies \omega_2 = 8\text{ rad/s}$.

68. Two parallel wires in the plane of the paper are distance $X_0$ apart. A point charge is moving with speed $u$ between the wires in the same plane at a distance $X_1$ from one of the wires. When the wires carry current of magnitude $I$ in the same direction, the radius of curvature of the path of the point charge is $R_1$. In contrast, if the currents $I$ in the two wires have directions opposite to each other, the radius of curvature of the path is $R_2$. If $\frac{X_0}{X_1} = 3$, the value of $\frac{R_1}{R_2}$ is [JEE Adv. 2014 P-I]

Answer: 3

Solution:

Substitute magnetic field expressions into $R = \frac{mv}{qB}$ for both cases to obtain the ratio.

69. A galvanometer gives full scale deflection with $0.006\text{ A}$ current. By connecting it to a $4990\ \Omega$ resistance, it can be converted into a voltmeter of range $0 - 30\text{ V}$. If connected to a $\frac{249}{n}\ \Omega$ resistance, it becomes an ammeter of range $0 - 1.5\text{ A}$. The value of $n$ is: [JEE Adv. 2014 P-I]

Answer: 5

Solution:

Using conversion formulas for galvanometer to voltmeter and ammeter yields $n = 5$.

70. In the xy-plane, the region $y > 0$ has a uniform magnetic field $B_1 \hat{k}$ and the region $y < 0$ has another uniform magnetic field $-B_2 \hat{k}$. A positively charged particle is projected from the origin along the positive y-axis with speed $v_0 = \pi\text{ ms}^{-1}$ at $t = 0$, as shown in the figure. Neglect gravity in this problem. Let $t = T$ be the time when the particle crosses the x-axis from below for the first time. If $\frac{B_2}{B_1} = 2$, the average speed of the particle, in $\text{ms}^{-1}$, along the x-axis in the time interval $t$ is ______. [JEE Adv. 2018 P-I]

Answer: 2

Solution:

Average speed along the x-axis is calculated using total displacement along x divided by total time $T$, giving $2\text{ ms}^{-1}$.

71. A moving coil galvanometer has $50$ turns and each turn has an area $2 \times 10^{-4}\text{ m}^2$. The magnetic field produced by the magnet inside the galvanometer is $0.02\text{ T}$. The torsional constant of the suspension wire is $10^{-4}\text{ Nm rad}^{-1}$. When a current flows through the galvanometer, a full scale deflection occurs if the coil rotates by $0.2\text{ rad}$. The resistance of the coil of the galvanometer is $50\ \Omega$. This galvanometer is to be converted into an ammeter capable of measuring current in the range $0 - 1.0\text{ A}$. For this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance, in ohms, is _______. [JEE Adv. 2018 P-II]

Answer: 6

Solution:

$$I_g = \frac{C\theta}{NABI} \implies I_g = 0.1\text{ A}$$ $$S = \frac{I_g R_g}{I - I_g} = \frac{0.1 \times 50}{1.0 - 0.1} = \frac{5}{0.9} \approx 5.56\ \Omega \approx 6\ \Omega$$

#####