JEE Physics Moving Charges and Magnetism MCQs Set 29

Answer: (c) 0.2 ohm

Explanation:
1. \( S=\dfrac{I_gG}{I-I_g}=\dfrac{0.002\times50}{0.498} \).
2. \( S\approx0.2\ \Omega \) in parallel.

Teacher's Note:
1. Write the force or torque law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (a) 5000

Explanation:
1. Magnetic damping force \( =\dfrac{B^2l^2}{R}v \), so \( b=\dfrac{0.01\times0.01}{10}=10^{-5} \) kg/s.
2. Amplitude falls by \( e \) at \( t=\dfrac{2m}b=10^4 \) s. Period \( T=2\pi\sqrt{m/k}=1.99 \) s, so \( N=\dfrac{10^4}{1.99}\approx5000 \).

Teacher's Note:
1. Write the force or torque law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (c) \( T_h=T_c \)

Explanation:
1. \( T=2\pi\sqrt{I/(\mu B)} \). Hoop: \( I=mR^2 \), moment \( 2\mu \). Cylinder: \( I=\tfrac12mR^2 \), moment \( \mu \).
2. \( I/\mu \) is \( \dfrac{mR^2}{2\mu} \) for both, so the periods are equal.

Teacher's Note:
1. Write the force or torque law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (a) \( \text{repulsive, }\dfrac{\mu_0I_1I_2}{4\pi} \)

Explanation:
1. Near side (distance \( a \)): \( F_1=\dfrac{\mu_0I_1I_2}{2\pi} \), repulsive (antiparallel currents).
2. Far side (distance \( 2a \)): \( F_2=\dfrac{\mu_0I_1I_2}{4\pi} \), attractive. Side forces cancel, so net \( =\dfrac{\mu_0I_1I_2}{4\pi} \) repulsive.

Teacher's Note:
1. Write the force or torque law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (c) \( \dfrac{2\mu_0Iqv}{\pi d} \)

Explanation:
1. At the midpoint each wire gives \( \dfrac{\mu_0I}{2\pi(d/2)}=\dfrac{\mu_0I}{\pi d} \), and with opposite currents the two fields add.
2. \( B=\dfrac{2\mu_0I}{\pi d} \), and as \( v\perp B \), \( F=qvB=\dfrac{2\mu_0Iqv}{\pi d} \).

Teacher's Note:
1. Write the force or torque law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (a) \( 2.56\times10^{-4}\ \text{Wb/m}^2 \)

Explanation:
1. O is on the equatorial line of each magnet at \( r=0.1 \) m: \( B=\dfrac{\mu_0}{4\pi}\dfrac m{r^3}=10^{-7}\dfrac m{10^{-3}} \), i.e. \( 1.2\times10^{-4} \) and \( 1.0\times10^{-4} \), both pointing North (opposite to the moments).
2. Adding Earth's field: \( 2.2\times10^{-4}+0.36\times10^{-4}=2.56\times10^{-4} \) Wb/m\(^2\).

Teacher's Note:
1. Write the force or torque law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (c) \( \dfrac{2NABi}{\pi} \)

Explanation:
1. Torque \( ki=NiAB \Rightarrow k=NAB \).
2. At equilibrium \( ki=C\theta \Rightarrow C=\dfrac{NABi}{\pi/2}=\dfrac{2NABi}{\pi} \).

Teacher's Note:
1. Write the force or torque law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (a) \( V_2=2V_1 \)

Explanation:
1. For identical geometry and current, \( V_H=\dfrac{BI}{ned}\propto\dfrac Bn \).
2. With \( B_1=B_2 \), \( \dfrac{V_2}{V_1}=\dfrac{n_1}{n_2}=2 \).

Teacher's Note:
1. Write the force or torque law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (c) \( B=5\ \text T,\ 75\ \text K\lt T_C(B)\lt 100\ \text K \)

Explanation:
1. \( T_C(B) \) decreases monotonically with B, so for \( 0\lt B\lt 7.5 \) T it lies between 75 K and 100 K, but its exact value is unknown.
2. For \( B=10 \) T it is below 75 K, so the third and fourth options fail; the first cannot be fixed to 80 K.

Teacher's Note:
1. Write the force or torque law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.