JEE Physics Moving Charges and Magnetism MCQs Set 23

Answer: (b) \( 2\,\mu T \) along \( +y \)

Explanation:
1. \( B=\frac{\mu_0 I}{2\pi r}=\frac{2\times10^{-7}\times10}{1}=2\times10^{-6}\,T = 2\,\mu T \).
2. Direction is along \( \hat I\times\hat r=\hat z\times\hat x=+\hat y \) (right-hand rule).
3. So the field is 2 microtesla along +y.

Teacher's Note:
1. Field of a long wire circles the wire; at a point on +x with current along +z it points along +y.
2. Use \( \mu_0/2\pi = 2\times10^{-7} \) T m/A to save time.

Answer: (a) \( 10^{4}\mu_0 \) m

Explanation:
1. \( B=\frac{\mu_0 I}{2\pi r}\Rightarrow r=\frac{\mu_0 I}{2\pi B} \).
2. \( r=\frac{\mu_0\,\pi}{2\pi\times5\times10^{-5}}=\frac{\mu_0}{10^{-4}}=10^{4}\mu_0 \) m.

Teacher's Note:
1. Keep \( \mu_0 \) symbolic when the options are written in terms of it.
2. Cancel π first to avoid arithmetic slips.

Answer: (d) one fourth

Explanation:
1. \( B=\frac{\mu_0 NI}{2R}\propto \frac IR \).
2. New value \( \propto \frac{I/2}{2R}=\frac14\cdot\frac IR \).

Teacher's Note:
1. Treat the quantities as ratios and combine the factors.
2. Both changes lower the field.

Answer: (d) four times

Explanation:
1. \( B=\mu_0\frac NL I \).
2. N to 2N and I to 2I gives \( B\to4B \).

Teacher's Note:
1. Both factors multiply, so the effects compound.
2. Do not forget that the length is fixed here.

Answer: (b) 1/2

Explanation:
1. \( r=\frac{p}{qB}\Rightarrow p=qBr \).
2. Same B and r give \( \frac{p_p}{p_\alpha}=\frac{q_p}{q_\alpha}=\frac{e}{2e}=\frac12 \).

Teacher's Note:
1. Radius depends on momentum per unit charge.
2. Alpha charge is 2e.

Answer: (a) \( 6.0\times10^{5} \) m/s

Explanation:
1. \( v=\frac{F}{qB}=\frac{3.84\times10^{-14}}{(3.2\times10^{-19})(0.2)} \).
2. \( v=6\times10^{5} \) m/s.

Teacher's Note:
1. The alpha particle has charge 2e = 3.2 × 10⁻¹⁹ C.
2. Check units: N / (C T) = m/s.

Answer: (d) \( \frac{24}{11} \)

Explanation:
1. \( r=\frac{mv}{qB}\propto\frac mq \).
2. \( \frac{r_A}{r_B}=\frac{24/1}{22/2}=\frac{24}{11} \).

Teacher's Note:
1. Same velocity and B means r ∝ m/q.
2. Be careful to divide each mass by its own charge.

Answer: (a) 1 : 2

Explanation:
1. \( r=\frac{mv}{qB}\propto\frac mq \) for the same v.
2. \( \frac{r_p}{r_\alpha}=\frac{1/1}{4/2}=\frac12 \).

Teacher's Note:
1. Same velocity gives r ∝ m/q.
2. Compare with the same-energy and same-potential cases, which give different ratios.

Answer: (c) E, B and v are mutually perpendicular and \( v=E/B \)

Explanation:
1. Undeflected motion needs \( qE=qvB \), with the two forces opposite.
2. That requires mutually perpendicular E, B and v with \( v=\frac EB \).

Teacher's Note:
1. This arrangement is the velocity selector.
2. The result does not depend on charge or mass.

Answer: (a) \( 2\times10^{-4} \) N, attractive

Explanation:
1. \( \frac Fl=\frac{\mu_0I_1I_2}{2\pi d}=\frac{2\times10^{-7}\times10\times10}{0.1}=2\times10^{-4} \) N/m.
2. Parallel currents attract.

Teacher's Note:
1. Currents in the same direction attract; opposite repel.
2. Convert 10 cm to 0.1 m.

Answer: (a) contract

Explanation:
1. Neighbouring turns carry current in the same direction.
2. Parallel currents attract, so the turns pull together and the spring contracts.

Teacher's Note:
1. Parallel currents attract.
2. This is the principle behind the 'mercury jumping spring' demonstration.

Answer: (b) \( \frac{2\pi BR^3}{\mu_0} \)

Explanation:
1. \( B=\frac{\mu_0I}{2R}\Rightarrow I=\frac{2BR}{\mu_0} \).
2. \( M=I\pi R^2=\frac{2BR}{\mu_0}\pi R^2=\frac{2\pi BR^3}{\mu_0} \).

Teacher's Note:
1. Eliminate I using the centre-field formula.
2. Check dimensions: M has units of A m².

Answer: (b) \( 10\times10^{16} \) rad/s

Explanation:
1. \( B=\frac{\mu_0I}{2R}\Rightarrow I=\frac{2RB}{\mu_0}=\frac{2\times10^{-10}\times16}{4\pi\times10^{-7}}\approx2.55\times10^{-3} \) A.
2. \( I=\frac{e\omega}{2\pi}\Rightarrow\omega=\frac{2\pi I}{e}=\frac{2\pi\times2.55\times10^{-3}}{1.6\times10^{-19}}\approx10^{17} \) rad/s.
3. This equals \( 10\times10^{16} \) rad/s.

Teacher's Note:
1. Find the equivalent current from the field, then use I = eω/2π.
2. 1 Å = 10⁻¹⁰ m.

Answer: (a) \( \frac MA \)

Explanation:
1. \( M=IA\Rightarrow I=\frac MA \).

Teacher's Note:
1. Magnetic moment is current × area for one turn.
2. Check units: A m² / m² = A.

Answer: (b) parallel to B

Explanation:
1. \( \tau=NIAB\sin\theta \) with \( \theta \) the angle between the area vector and B.
2. Maximum at \( \theta=90^\circ \), i.e. when the plane of the coil is parallel to B.

Teacher's Note:
1. Angle between normal and field, not between plane and field.
2. Plane perpendicular to B means zero torque.

Answer: (a) maximum

Explanation:
1. \( \tau=MB\sin\theta \) with \( \theta=90^\circ \) between M and B when the plane is parallel to B.
2. So the torque is maximum.

Teacher's Note:
1. Parallel to the plane means perpendicular to the area vector.
2. The loop tends to rotate until its plane is perpendicular to B.

Answer: (a) its speed remains unchanged

Explanation:
1. The force is perpendicular to v, so it changes the direction but does no work.
2. Speed is constant; velocity and acceleration (directions) change continuously.

Teacher's Note:
1. Velocity is a vector, speed is a magnitude.
2. Uniform circular motion has constant speed but changing velocity.

Answer: (b) [L²A]

Explanation:
1. \( M=IA \): dimensions of current times area, \( [A][L^2] \).

Teacher's Note:
1. The unit A m² gives the dimensions directly.
2. No time dimension appears.

Answer: (c) in a circular orbit

Explanation:
1. With only B acting perpendicular to v, the force is always perpendicular to the velocity and constant in magnitude.
2. That produces uniform circular motion.

Teacher's Note:
1. Constant speed with a perpendicular force gives a circle.
2. A parabola would need a constant-direction force.

Answer: (a) T

Explanation:
1. Magnetic force is perpendicular to velocity and does no work.
2. So the speed and kinetic energy remain T at all times.

Teacher's Note:
1. Time is a distractor.
2. Only an electric field could change KE.

Answer: (a) \( \frac{\mu_0qf}{2R} \)

Explanation:
1. Equivalent current \( I=qf \).
2. \( B=\frac{\mu_0I}{2R}=\frac{\mu_0qf}{2R} \).

Teacher's Note:
1. A rotating charged ring is a current loop.
2. I = charge per revolution × revolutions per second.

Answer: (b) \( 1.5\times10^{-2} \) N m

Explanation:
1. \( M=NIA=2000\times2\times1.5\times10^{-4}=0.6 \) A m².
2. \( \tau=MB\sin30^\circ=0.6\times5\times10^{-2}\times0.5=1.5\times10^{-2} \) N m.

Teacher's Note:
1. The solenoid acts as a bar magnet along its axis.
2. Use the angle between M (axis) and B.