JEE Physics Moving Charges and Magnetism MCQs Set 39

Answer: (c) (2) and (3) only

Explanation:
1. Drift speed \( v_d=\dfrac{I}{newd} \) and the Hall field \( E=v_dB \) acts across the thickness d.
2. Voltage \( V=Ed=\dfrac{BI}{new} \): it depends on the width w but not on the thickness d.
3. So (2) is true (d does not matter), and halving the width doubles V, so (3) is true; (1) and (4) are false.

Teacher's Note:
1. The Hall voltage does not depend on the dimension along which it is measured.
2. Narrower strips give a larger Hall voltage.

Answer: (b) (1), (3) and (4) only

Explanation:
1. For equal dimensions and current, \( V=\dfrac{BI}{new}\propto\dfrac{B}{n} \).
2. (1): \( V_2/V_1=n_1/n_2=2 \), true, so (2) is false. (3): \( V_2/V_1=\dfrac{B_2}{B_1}=\dfrac12 \), true.
3. (4): \( \dfrac{V_2}{V_1}=\dfrac{B_2}{B_1}\cdot\dfrac{n_1}{n_2}=\dfrac12\times2=1 \), true.

Teacher's Note:
1. Hall voltage varies directly as B and inversely as the carrier density.
2. Take the ratio of the two voltages.

Answer: (a) (2) and (3) only

Explanation:
1. Voltage range: all in series gives \( V_1=I_g(2R+2R_C)<3RI_g \). Option (2) gives \( V_2=2I_g\left(2R+\dfrac{R_C}{2}\right)=I_g(4R+R_C)>4RI_g \), which is larger, so (2) is true and (1) is false.
2. Current range: all in parallel gives \( I=2I_g\left(1+\dfrac{R_C}{R}\right) \), while option (4) gives \( I=I_g\left(1+\dfrac{4R_C}{R}\right) \).
3. The difference \( (1-2R_C/R)I_g>0 \) because \( R_C<R/2 \), so (3) is larger and true; (4) is false.

Teacher's Note:
1. Compare the actual full-scale voltage or current of each arrangement.
2. The condition \( R_C<R/2 \) decides the comparison.

Answer: (d) 5.56 ohm

Explanation:
1. Full-scale current: \( I_g=\dfrac{C\theta}{NAB}=\dfrac{10^{-4}\times0.2}{50\times2\times10^{-4}\times0.02}=0.1 \) A.
2. Shunt: \( I_gG=(I-I_g)S \), so \( S=\dfrac{0.1\times50}{0.9}=5.56 \) ohm.

Teacher's Note:
1. First find the current that gives full-scale deflection.
2. The voltage across the galvanometer and the shunt is the same.

Answer: (a) \( \dfrac{2\pi m}{qB} \)

Explanation:
1. Magnetic force provides the centripetal force: \( qvB=\dfrac{mv^2}{r} \).
2. Period \( T=\dfrac{2\pi r}{v}=\dfrac{2\pi m}{qB} \).

Teacher's Note:
1. The period of circular motion does not depend on the speed or radius.
2. Write the force balance first.

Answer: (c) a force and a torque

Explanation:
1. A needle is a dipole with two poles. In a non-uniform field the forces on the two poles are unequal, so there is a net force.
2. The forces are not along the same line unless the needle is aligned with the field, so there is also a torque.

Teacher's Note:
1. In a uniform field a dipole feels only a torque.
2. Non-uniformity adds a net force.

Answer: (d) \( 5\times10^{-5} \)

Explanation:
1. Field of each coil: \( B=\dfrac{\mu_0I}{2R}=\dfrac{4\pi\times10^{-7}I}{2\times2\pi\times10^{-2}}=10^{-5}I \).
2. So \( B_1=3\times10^{-5} \) and \( B_2=4\times10^{-5} \) and they are perpendicular.
3. Resultant \( \sqrt{3^2+4^2}\times10^{-5}=5\times10^{-5} \).

Teacher's Note:
1. Coils at right angles have perpendicular axes, so the fields add as vectors.
2. Simplify the coil formula with \( R=2\pi \) cm.

Answer: (b) attract each other with a force per unit length of \( \dfrac{\mu_0I^2}{2\pi d} \)

Explanation:
1. Parallel currents attract.
2. Field of one wire at the other: \( B=\dfrac{\mu_0I}{2\pi d} \), so force per unit length \( =BI=\dfrac{\mu_0I^2}{2\pi d} \).

Teacher's Note:
1. The force per unit length falls off as the first power of the distance.
2. Currents in the same direction attract.

Answer: (a) \( 9995 \) ohm

Explanation:
1. One division = 0.1 mA = 0.5 mV, so the coil resistance is \( G=\dfrac{0.5\ \text{mV}}{0.1\ \text{mA}}=5 \) ohm.
2. For 1 V per division at 0.1 mA, total resistance \( =\dfrac{1}{10^{-4}}=10^4 \) ohm.
3. Series resistance \( 10^4-5=9995 \) ohm.

Teacher's Note:
1. Find the coil resistance from the two sensitivities.
2. Subtract it from the required total.

Answer: (b) straight line

Explanation:
1. The electric force accelerates the particle along the field direction.
2. Its velocity is then parallel to B, so the magnetic force \( q\vec v\times\vec B \) is zero. The particle moves in a straight line.

Teacher's Note:
1. Velocity along B means no magnetic force.
2. Starting from rest the velocity is along E.

Answer: (d) the momentum changes but the kinetic energy is constant

Explanation:
1. The magnetic force is perpendicular to the velocity, so it changes the direction of velocity (hence momentum) but does no work.
2. Speed and kinetic energy remain constant.

Teacher's Note:
1. Momentum is a vector and its direction changes.
2. No work means no change in kinetic energy.

Answer: (d) the magnetic field at any point inside the pipe is zero

Explanation:
1. Take an Amperian loop inside the pipe. It encloses no current, so by Ampere's law \( \oint\vec B\cdot d\vec l=0 \).
2. By symmetry the field is tangential and uniform on the loop, so it is zero everywhere inside.

Teacher's Note:
1. Apply Ampere's law to a circle inside the pipe.
2. No enclosed current means zero field by symmetry.

Answer: (c) 1

Explanation:
1. Inside: \( B=\dfrac{\mu_0Ir}{2\pi a^2} \) so at \( a/2 \): \( \dfrac{\mu_0I}{4\pi a} \).
2. Outside: \( B=\dfrac{\mu_0I}{2\pi r} \) so at \( 2a \): \( \dfrac{\mu_0I}{4\pi a} \).
3. The ratio is 1.

Teacher's Note:
1. Use Ampere's law inside and outside separately.
2. Inside the field grows linearly with r.

Answer: (a) \( \dfrac{\mu_0}{2\pi d}\left(I_1^2+I_2^2\right)^{1/2} \)

Explanation:
1. The point is at distance d from both wires. Field magnitudes: \( \dfrac{\mu_0I_1}{2\pi d} \) and \( \dfrac{\mu_0I_2}{2\pi d} \).
2. The two fields are perpendicular, so \( B=\dfrac{\mu_0}{2\pi d}\sqrt{I_1^2+I_2^2} \).

Teacher's Note:
1. Directions of the two fields are at right angles.
2. Add perpendicular vectors in quadrature.

Answer: (d) \( 5\times10^{-6} \) T, southward

Explanation:
1. Magnitude \( B=\dfrac{\mu_0I}{2\pi r}=\dfrac{2\times10^{-7}\times100}{4}=5\times10^{-6} \) T.
2. Direction by the right-hand rule: the current flows towards the west and the point is below the wire, so the field points to the south.

Teacher's Note:
1. Use the right-hand thumb rule for direction.
2. Take the distance as the height of the wire.

Answer: (a) \( \varepsilon_r=1.5,\ \mu_r=0.5 \)

Explanation:
1. Any dielectric has \( \varepsilon_r>1 \).
2. A diamagnetic material has \( \mu_r<1 \).
3. Only \( \varepsilon_r=1.5,\mu_r=0.5 \) satisfies both.

Teacher's Note:
1. Relative permittivity of matter is always more than 1.
2. Diamagnets have permeability slightly less than that of vacuum.

Answer: (c) \( \dfrac{\mu_0I(b-a)}{24ab} \)

Explanation:
1. The radial segments pass through O and give no field.
2. Arc of angle \( \theta=\pi/6 \): \( B=\dfrac{\mu_0I\theta}{4\pi r} \). The two arcs carry current in opposite senses, so the fields subtract.
3. Net \( B=\dfrac{\mu_0I}{4\pi}\cdot\dfrac\pi6\left(\dfrac1a-\dfrac1b\right)=\dfrac{\mu_0I(b-a)}{24ab} \).

Teacher's Note:
1. Straight segments along a radius give zero field at the centre.
2. Currents in the two arcs circulate in opposite directions.

Answer: (b) \( I_1=I_2=\dfrac{B\ell v}{3R},\ I=\dfrac{2B\ell v}{3R} \)

Explanation:
1. PQ is the source of emf \( B\ell v \) with internal resistance R. The two arms are in parallel: \( R/2 \).
2. Current \( I=\dfrac{B\ell v}{R+R/2}=\dfrac{2B\ell v}{3R} \).
3. It splits equally: \( I_1=I_2=\dfrac{B\ell v}{3R} \).

Teacher's Note:
1. Treat the moving rod as a battery.
2. Equal parallel branches share the current equally.

Answer: (b) \( \dfrac{\mu_0I}{\pi^2R} \)

Explanation:
1. Divide the ring into thin strips subtending \( d\theta \), each carrying current \( dI=\dfrac{I}{\pi}d\theta \).
2. Each gives \( dB=\dfrac{\mu_0dI}{2\pi R} \) at the axis, perpendicular to the radius. Components along the symmetry axis add up while the perpendicular components cancel.
3. Net \( B=\int_0^\pi\dfrac{\mu_0I}{2\pi^2R}\sin\theta\,d\theta=\dfrac{\mu_0I}{\pi^2R} \).

Teacher's Note:
1. Break the half-cylinder into thin wires and integrate.
2. Use symmetry to cancel one component.

Answer: (d) 0.15 mV

Explanation:
1. \( \varepsilon=B\ell v=5.0\times10^{-5}\times2\times1.5=1.5\times10^{-4} \) V = 0.15 mV.

Teacher's Note:
1. The boat velocity, field and aerial are mutually perpendicular.
2. Motional emf is \( B\ell v \).

Answer: (c) \( B\propto\dfrac1R \)

Explanation:
1. Ring of radius r: \( dI=\dfrac{Q}{\pi R^2}2\pi r\,dr\,\dfrac{\omega}{2\pi}=\dfrac{Q\omega r\,dr}{\pi R^2} \).
2. Field at the centre \( dB=\dfrac{\mu_0dI}{2r} \) so \( B=\dfrac{\mu_0Q\omega}{2\pi R^2}\int_0^Rdr=\dfrac{\mu_0Q\omega}{2\pi R} \).
3. So B varies inversely as R.

Teacher's Note:
1. Use the surface density \( Q/\pi R^2 \) and integrate over rings.
2. With Q fixed, the density falls as \( R^2 \).

Answer: (b) \( r_\alpha=r_p<r_d \)

Explanation:
1. For the same kinetic energy K: \( r=\dfrac{mv}{qB}=\dfrac{\sqrt{2mK}}{qB}\propto\dfrac{\sqrt m}{q} \).
2. Proton: \( \sqrt{1}/1=1 \). Deuteron: \( \sqrt2 \). Alpha: \( \sqrt4/2=1 \).
3. So \( r_\alpha=r_p<r_d \).

Teacher's Note:
1. Express the radius in terms of kinetic energy.
2. Mass ratios are 1 : 2 : 4 and charge ratios 1 : 1 : 2.

Answer: (d) 3 A

Explanation:
1. Field intensity in the solenoid \( H=nI=\dfrac{N}{L}I \) with \( n=\dfrac{100}{0.1}=1000 \) per metre.
2. Set \( H=3\times10^3 \): \( I=\dfrac{3000}{1000}=3 \) A.

Teacher's Note:
1. Coercivity is a magnetising field in A/m, so compare it with \( nI \).
2. Convert length to metres.

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

Explanation:
1. Separation of the wires: \( d=2L\sin\theta \). Magnetic force per unit length \( F=\dfrac{\mu_0I^2}{2\pi d} \).
2. Equilibrium of a unit length: \( \tan\theta=\dfrac{F}{\lambda g} \).
3. So \( I^2=\dfrac{2\pi d\lambda g\tan\theta}{\mu_0}=\dfrac{4\pi L\lambda g\sin\theta\tan\theta}{\mu_0} \), giving \( I=2\sin\theta\sqrt{\dfrac{\pi\lambda gL}{\mu_0\cos\theta}} \).

Teacher's Note:
1. Resolve the tension into vertical and horizontal components.
2. Find the separation from the geometry of the threads.

Answer: (d) \( \dfrac{\pi^2}{8\sqrt2} \)

Explanation:
1. Circle: \( l=2\pi R \), \( B_A=\dfrac{\mu_0I}{2R}=\dfrac{\pi\mu_0I}{l} \).
2. Square: \( a=\dfrac l4 \), \( B_B=\dfrac{2\sqrt2\mu_0I}{\pi a}=\dfrac{8\sqrt2\mu_0I}{\pi l} \).
3. Ratio \( \dfrac{\pi}{8\sqrt2/\pi}=\dfrac{\pi^2}{8\sqrt2} \).

Teacher's Note:
1. Express both fields in terms of the wire length l.
2. Square field at the centre is \( 2\sqrt2\mu_0I/\pi a \).

Answer: (b) \( \sqrt2 \)

Explanation:
1. Moment \( m=I\pi R^2 \), so doubling m at constant I doubles the area and \( R\to\sqrt2R \).
2. Field \( B=\dfrac{\mu_0I}{2R}\propto\dfrac1R \), so \( \dfrac{B_1}{B_2}=\sqrt2 \).

Teacher's Note:
1. At constant current, moment is proportional to the area.
2. Field at the centre varies inversely with the radius.

Answer: (c) \( r_e<r_p=r_\alpha \)

Explanation:
1. \( r=\dfrac{\sqrt{2mK}}{qB}\propto\dfrac{\sqrt m}{q} \).
2. Electron: smallest because its mass is smallest. Proton: \( \sqrt{m_p} \). Alpha: \( \dfrac{\sqrt{4m_p}}{2}=\sqrt{m_p} \).

Teacher's Note:
1. Same kinetic energy: compare \( \sqrt m/q \).
2. The electron's mass is about 1/1836 of the proton's.