JEE Physics Moving Charges and Magnetism MCQs Set 13

Answer: (a) \( IBR \)

Explanation:
1. The magnetic force on the wire equals that on the straight chord: \( F=IB(2R\sin\theta_0) \), directed radially outwards.
2. The two end tensions balance it: \( 2T\sin\theta_0=2IBR\sin\theta_0\Rightarrow T=IBR \)

Teacher's Note:
1. A small element is in equilibrium under tension and the outward magnetic force per unit length IB, like a rope held by pressure.

Answer: (a) \( \vec B\perp\vec v \)

Explanation:
1. \( \vec B=\frac{\mu_0}{4\pi}\frac{q\,\vec v\times\hat r}{r^2} \) shows that B is perpendicular to both v and the line joining the charge to the point.
2. It falls as the inverse square of the distance, so the other options are wrong.

Teacher's Note:
1. The cross product in the Biot-Savart law makes B perpendicular to v.

Answer: (b) \( -\frac{2F}{3} \)

Explanation:
1. \( F\propto\frac{I_1I_2}{d} \)
2. New force: \( F'=\frac{(2I_1)(-I_2)}{3d}\cdot k=-\frac23F \)

Teacher's Note:
1. Force between parallel wires is proportional to I1I2/d; reversing a current reverses the sign.

Answer: (d) \( \frac{Bv_0}{E} \)

Explanation:
1. In E: \( \frac{mv_0^2}{r_1}=eE\Rightarrow r_1=\frac{mv_0^2}{eE} \).
2. In B: \( r_2=\frac{mv_0}{eB} \).
3. \( \frac{r_1}{r_2}=\frac{Bv_0}{E} \)

Teacher's Note:
1. Radius of curvature is mv²/(force component perpendicular to the velocity).

Answer: (d) The electron will continue to move with uniform velocity along the axis of the solenoid

Explanation:
1. Inside a long solenoid the field is uniform and parallel to the axis.
2. The force \( q\vec v\times\vec B=0 \) when v is parallel to B, so the velocity is unchanged.

Teacher's Note:
1. A charge moving parallel to a magnetic field feels no magnetic force.

Answer: (a) 1

Explanation:
1. \( B=\frac{\mu_0I}{2R}:\ B_A=\frac{\mu_0I}{2R},\quad B_B=\frac{\mu_0(2I)}{2(2R)}=\frac{\mu_0I}{2R} \)

Teacher's Note:
1. The field at the centre of a loop is proportional to I/R.

Answer: (b) \( n^2B \)

Explanation:
1. Radius becomes R/n (same length of wire).
2. \( B'=\frac{\mu_0nI}{2(R/n)}=n^2\frac{\mu_0I}{2R}=n^2B \)

Teacher's Note:
1. Both the number of turns and the 1/radius factor increase by n.

Answer: (a) Undergoes acceleration all the time

Explanation:
1. Inside a dee the particle moves on a circle at constant speed, which is still an acceleration (centripetal).
2. Between the dees the electric field increases its speed, so it accelerates at every stage.

Teacher's Note:
1. Speeding up is due to the electric field in the gap; the magnetic field only bends the path.

Answer: (c) A force and a torque

Explanation:
1. The two poles experience unequal forces in a non-uniform field, giving a net force.
2. The needle can also be inclined to the field, giving a torque.

Teacher's Note:
1. Torque depends on orientation; net force needs a field gradient.

Answer: (d) \( 2\sin\theta\sqrt{\frac{\pi\lambda gL}{\mu_0\cos\theta}} \)

Explanation:
1. Separation \( d=2L\sin\theta \); repulsive force per unit length \( \frac{\mu_0I^2}{2\pi d} \).
2. Equilibrium: \( \tan\theta=\frac{\mu_0I^2/(4\pi L\sin\theta)}{\lambda g} \).
3. \( I=2\sqrt{\frac{\pi\lambda gL\sin\theta\tan\theta}{\mu_0}}=2\sin\theta\sqrt{\frac{\pi\lambda gL}{\mu_0\cos\theta}} \)

Teacher's Note:
1. Resolve the thread tension into vertical (weight) and horizontal (magnetic repulsion) parts.

Answer: (b) Zero

Explanation:
1. The magnetic force is always perpendicular to the velocity, so it does no work on the particle at any instant.
2. Hence the work in a full circle is zero.

Teacher's Note:
1. A force perpendicular to displacement does no work, so a magnetic field cannot change kinetic energy.

Answer: (a) Current sensitivity of the galvanometer is 20 μA/division

Explanation:
1. Deflection is proportional to current: \( \frac{2}{2400+G}=40k,\quad\frac{2}{4900+G}=20k \).
2. Dividing: \( 4900+G=2(2400+G)\Rightarrow G=100\ \Omega \).
3. Current for 40 divisions \( =\frac{2}{2500}=0.8\ \text{mA} \), so sensitivity is 20 μA/division; full scale (50 divisions) is 1 mA; for 10 divisions R = 9900 Ω.

Teacher's Note:
1. Use the two readings to eliminate the unknown sensitivity and find G.

Answer: (d) rs = 0.5 Ω in parallel with the galvanometer

Explanation:
1. An ammeter must have very low resistance, so rs must be in parallel.
2. Ammeter resistance \( =\frac{100\times0.5}{100.5}=0.4975\ \Omega \) gives current \( \frac{5}{50.4975}=0.0990\ \text{A} \), 0.99% below 0.1 A (within 1%).
3. With 1 Ω the resistance is 0.99 Ω and the current falls by 1.9%, outside the limit.

Teacher's Note:
1. Compute the fractional change in current caused by the added ammeter resistance.

Answer: (d) \( 10^{3}\ \text{Wb/m}^2 \)

Explanation:
1. Undeflected motion needs \( qE=qvB \).
2. \( B=\frac Ev=\frac{10^4}{10}=10^3\ \text{Wb/m}^2 \)

Teacher's Note:
1. In a velocity selector the magnitude of B is E/v; the charge cancels.