JEE Physics Moving Charges and Magnetism MCQs Set 38

Answer: (a) zero

Explanation:
1. The midpoint is at the same distance from both coils, so the two coils produce fields of equal magnitude there.
2. The currents are opposite, so the two fields point in opposite directions along the axis and cancel.

Teacher's Note:
1. Check the symmetry of the point before calculating anything.
2. Opposite currents in identical coils give opposite axial fields.

Answer: (c) \( 1.3\times10^{-4} \) T

Explanation:
1. Own coil at its centre: \( B_1=\dfrac{\mu_0NI}{2R}=\dfrac{4\pi\times10^{-7}\times10\times1}{2\times0.04}=1.57\times10^{-4} \) T.
2. Other coil on the axis at \( x=6 \) cm: \( B_2=\dfrac{\mu_0NIR^2}{2(R^2+x^2)^{3/2}}=2.7\times10^{-5} \) T, in the opposite direction.
3. Net \( B=1.57\times10^{-4}-0.27\times10^{-4}\approx1.3\times10^{-4} \) T.

Teacher's Note:
1. Subtract the fields because the currents are opposite.
2. Use the on-axis formula for the second coil.

Answer: (d) 1 : 4

Explanation:
1. Length \( L=2\pi r \) is fixed. For two turns the radius becomes \( r/2 \).
2. One turn: \( B_1=\dfrac{\mu_0I}{2r} \). Two turns: \( B_2=\dfrac{2\mu_0I}{2(r/2)}=\dfrac{2\mu_0I}{r} \).
3. So \( B_1:B_2=1:4 \).

Teacher's Note:
1. The same wire length means the radius halves when the turns double.
2. Field at the centre is \( N\mu_0I/2R \).

Answer: (a) \( \dfrac{Bid}{nAe} \)

Explanation:
1. Drift velocity \( v_d=\dfrac{i}{nAe} \).
2. Transverse field \( E=v_dB=\dfrac{Bi}{nAe} \) balances the magnetic force on the electrons.
3. Hall voltage \( V=Ed=\dfrac{Bid}{nAe} \).

Teacher's Note:
1. The Hall field stops further accumulation of charge when the electric and magnetic forces on a carrier are equal.
2. Voltage is field times the width across which it acts.

Answer: (d) 3.87 C

Explanation:
1. Speed gained: \( v=\sqrt{2gh}=\sqrt{60}=7.75 \) m/s.
2. Impulse of the magnetic force: \( \int F\,dt=B\ell\int i\,dt=B\ell q=mv \).
3. So \( q=\dfrac{mv}{B\ell}=\dfrac{0.01\times7.75}{0.1\times0.2}=3.87 \) C.

Teacher's Note:
1. Impulse equals the change in momentum, and the force on the wire is \( B i \ell \).
2. Convert grams and centimetres to SI units first.

Answer: (b) \( 2.8\times10^{-5} \) J/m

Explanation:
1. Force per unit length \( \dfrac{F}{\ell}=\dfrac{\mu_0i_1i_2}{2\pi r} \) (attractive).
2. Work per unit length \( =\int_{r_1}^{r_2}\dfrac{\mu_0i_1i_2}{2\pi r}dr=\dfrac{\mu_0i_1i_2}{2\pi}\ln\dfrac{r_2}{r_1} \).
3. Value: \( 2\times10^{-7}\times200\times\ln2=2.77\times10^{-5} \) J/m.

Teacher's Note:
1. The force changes with distance, so integrate.
2. Natural logarithm of the ratio of separations appears.

Answer: (d) \( \dfrac{3mv^2}{4qa} \)

Explanation:
1. The magnetic force does no work. Only the electric force does work, through the displacement \( 2a \) along x.
2. Work-energy theorem: \( qE(2a)=\tfrac12m(2v)^2-\tfrac12mv^2=\tfrac32mv^2 \).
3. So \( E=\dfrac{3mv^2}{4qa} \).

Teacher's Note:
1. A magnetic force is always perpendicular to velocity, so it does zero work.
2. The electric force is along x, so only the x-displacement matters.

Answer: (c) \( \dfrac{3mv^3}{4a} \)

Explanation:
1. From the work-energy theorem \( E=\dfrac{3mv^2}{4qa} \).
2. Power \( P=q\vec E\cdot\vec v=qEv=\dfrac{3mv^3}{4a} \).

Teacher's Note:
1. Power is force dotted with velocity.
2. Here the electric force and the velocity are parallel.

Answer: (b) \( \dfrac{m}{qB}\sin^{-1}\!\left(dB\sqrt{\dfrac{q}{2mV}}\right) \)

Explanation:
1. Speed \( v=\sqrt{2qV/m} \) and radius \( r=\dfrac{mv}{qB} \).
2. The particle turns through angle \( \theta \) with \( \sin\theta=\dfrac{d}{r}=\dfrac{dqB}{mv}=dB\sqrt{\dfrac{q}{2mV}} \).
3. Time \( t=\dfrac{\theta}{\omega}=\dfrac{m\theta}{qB} \).

Teacher's Note:
1. Geometry: the width of the region is the sine component of the radius.
2. Time is angle turned divided by angular frequency \( qB/m \).

Answer: (c) 30 degrees

Explanation:
1. \( I\propto\tan\theta \). At 45 degrees \( \tan\theta=1 \).
2. New current is \( I/\sqrt3 \), so \( \tan\theta'=1/\sqrt3 \) giving \( \theta'=30^\circ \).

Teacher's Note:
1. A tangent galvanometer has current proportional to tan of the deflection.
2. Do not scale the angle directly.

Answer: (d) \( \dfrac{\mu_0II'c(b-a)}{2\pi ab} \) towards the wire

Explanation:
1. The horizontal sides have equal and opposite forces and cancel.
2. Near side (current parallel to I): attracted, \( F_1=\dfrac{\mu_0II'c}{2\pi a} \). Far side (antiparallel): repelled, \( F_2=\dfrac{\mu_0II'c}{2\pi b} \).
3. Net \( F_1-F_2=\dfrac{\mu_0II'c}{2\pi}\left(\dfrac1a-\dfrac1b\right) \), towards the wire.

Teacher's Note:
1. Parallel currents attract and antiparallel currents repel.
2. The nearer side wins because the force falls with distance.

Answer: (c) \( 5.9\times10^{-2} \) N m

Explanation:
1. Moment \( m=NIA=6\times2\times\pi(0.15)^2=0.85 \) A m².
2. The field lies in the plane of the clock, so it is perpendicular to the moment and \( \sin\theta=1 \).
3. Torque \( \tau=mB=0.85\times0.07\approx5.9\times10^{-2} \) N m.

Teacher's Note:
1. Torque is \( NIAB\sin\theta \).
2. A field in the plane of a loop is perpendicular to its axis.

Answer: (a) \( 4.7\times10^{-3} \) T

Explanation:
1. Speed \( v=\sqrt{2K/m}=2.65\times10^7 \) m/s. The component along B is \( v\cos60^\circ \).
2. The electron moves on a helix about GS and meets the axis again after one period \( T=\dfrac{2\pi m}{eB} \).
3. For the minimum B, \( GS=v\cos60^\circ\cdot\dfrac{2\pi m}{eB} \), so \( B=\dfrac{2\pi mv\cos60^\circ}{e\cdot GS}\approx4.7\times10^{-3} \) T.

Teacher's Note:
1. Resolve the velocity along and perpendicular to the field.
2. The smallest B corresponds to exactly one turn of the helix.

Answer: (c) \( \dfrac{\mu_0J}{2}\left(\dfrac R2-\dfrac{a^2}{2R-b}\right) \)

Explanation:
1. Use superposition: a full cylinder of current density \( J \) plus a smaller cylinder (the hole) of current density \( -J \).
2. Full cylinder at distance \( 2R \): \( B_1=\dfrac{\mu_0J\pi R^2}{2\pi(2R)}=\dfrac{\mu_0JR}{4} \).
3. The hole's axis is at distance \( 2R-b \): \( B_2=\dfrac{\mu_0J\pi a^2}{2\pi(2R-b)} \) in the opposite sense.
4. Net \( B=\dfrac{\mu_0J}{2}\left(\dfrac R2-\dfrac{a^2}{2R-b}\right) \).

Teacher's Note:
1. Treat the hole as a cylinder carrying the opposite current density.
2. Both points lie on the same line, so the two fields are collinear.

Answer: (a) \( \dfrac{NABQ}{\sqrt{IC}} \)

Explanation:
1. Angular impulse: \( \int NAB\,i\,dt=NABQ=I\omega_0 \), so \( \omega_0=\dfrac{NABQ}{I} \).
2. Kinetic energy turns into twist energy: \( \tfrac12I\omega_0^2=\tfrac12C\theta_m^2 \).
3. Hence \( \theta_m=\omega_0\sqrt{\dfrac IC}=\dfrac{NABQ}{\sqrt{IC}} \).

Teacher's Note:
1. Treat the charge pulse as an impulse that gives the coil an initial angular velocity.
2. Use energy conservation for the swing.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. Assertion is true: a soft iron core concentrates the field lines, increasing the field through the coil and hence the sensitivity.
2. Reason is false: soft iron has high permeability but it is easily magnetised and demagnetised (low retentivity and low coercivity).

Teacher's Note:
1. Soft iron: high permeability, low coercivity, easy to magnetise and demagnetise.
2. That is why it is used for electromagnet and transformer cores.

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

Explanation:
1. Radius of the path \( r=\dfrac{mv}{qB} \). The particle crosses into region III only if \( r>l \), i.e. \( v>\dfrac{qlB}{m} \). So (1) is true and (2) is false.
2. If it returns, it describes a semicircle, length \( \pi r \), which grows with v and is greatest when \( r=l \) i.e. \( v=\dfrac{qlB}{m} \). So (3) is true.
3. Time for the semicircle is \( \dfrac{\pi m}{qB} \), independent of v. So (4) is true.

Teacher's Note:
1. For motion in a field, the radius grows with speed but the period does not.
2. Compare the radius with the width of the region.

Answer: (a) 7

Explanation:
1. Sides PQ and PR pass through P, so they produce no field at P.
2. Distance of P from QR: \( d=\dfrac{3x\times4x}{5x}=\dfrac{12x}{5} \). The angles subtended give \( \sin\alpha=\dfrac35 \) and \( \sin\beta=\dfrac45 \).
3. Field \( B=\dfrac{\mu_0I}{4\pi d}(\sin\alpha+\sin\beta)=\dfrac{\mu_0I}{4\pi}\cdot\dfrac{5}{12x}\cdot\dfrac75=\dfrac{7\mu_0I}{48\pi x} \), so k = 7.

Teacher's Note:
1. A straight wire gives no field on its own line.
2. Use \( \dfrac{\mu_0I}{4\pi d}(\sin\alpha+\sin\beta) \) for the third side.

Answer: (d) \( B=5 \) T, \( 75\ \text{K}<T_c(B)<100\ \text{K} \)

Explanation:
1. The critical temperature falls as B rises. At 5 T (less than 7.5 T) it lies between 75 K and 100 K, but its exact value is unknown, so 80 K cannot be asserted.
2. At 10 T (more than 7.5 T) it is below 75 K, so the range 75 K to 100 K is impossible and 70 K is only possible, not certain.

Teacher's Note:
1. Use the monotonic fall of critical temperature with field.
2. 'Definitely' means true for every possible curve shape.

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

Explanation:
1. Radius of the circle: \( r=\dfrac{l}{2\pi} \).
2. For a small element, \( 2T\sin\dfrac{d\theta}{2}\approx T\,d\theta=I\,(r\,d\theta)\,B \) so \( T=IBr \).
3. Hence \( T=\dfrac{IBl}{2\pi} \).

Teacher's Note:
1. Balance the outward magnetic force on a small arc with the tension components.
2. The magnetic force on an element is \( IB\,dl \).

Answer: (d) \( \dfrac{\mu_0NI}{2(b-a)}\ln\dfrac ba \)

Explanation:
1. Number of turns in a thin ring of radius between \( r \) and \( r+dr \): \( dn=\dfrac{N}{b-a}dr \).
2. Its field at the centre: \( dB=\dfrac{\mu_0I\,dn}{2r}=\dfrac{\mu_0NI}{2(b-a)}\dfrac{dr}{r} \).
3. Integrating from a to b: \( B=\dfrac{\mu_0NI}{2(b-a)}\ln\dfrac ba \).

Teacher's Note:
1. Treat the spiral as many thin concentric rings.
2. The integral of \( dr/r \) gives a logarithm.

Answer: (a) (3) and (4) only

Explanation:
1. E and B are both along y. The electric force \( qE_0 \) along y keeps increasing the y-component of velocity, while the magnetic force only rotates the velocity components in the x-z plane.
2. Hence the motion is a helix about the y-axis with increasing pitch whenever there is a perpendicular component of velocity. For \( \theta=0^\circ \) it is not a plane circle (it drifts along y) and the pitch is not constant, so (1) and (2) are false; for \( \theta=10^\circ \) (3) is true.
3. For \( \theta=90^\circ \) the velocity is along B, so the magnetic force is zero and the charge accelerates linearly along y: (4) is true.

Teacher's Note:
1. Resolve the motion along and perpendicular to B.
2. A force along the axis of the helix makes the pitch grow.

Answer: (c) 5

Explanation:
1. Full cylinder (radius a) at distance a from its axis: \( B_1=\dfrac{\mu_0J\pi a^2}{2\pi a}=\dfrac{\mu_0Ja}{2} \).
2. Cavity (radius \( a/2 \)) at distance \( 3a/2 \) from P with current density \( -J \): \( B_2=\dfrac{\mu_0J\pi(a/2)^2}{2\pi(3a/2)}=\dfrac{\mu_0Ja}{12} \).
3. Net \( B=\dfrac{\mu_0Ja}{2}-\dfrac{\mu_0Ja}{12}=\dfrac{5}{12}\mu_0aJ \), so N = 5.

Teacher's Note:
1. Use superposition with a negative current density in the cavity.
2. Both fields are in the same line, so subtract.

Answer: (b) \( 2.56\times10^{-4} \) Wb/m²

Explanation:
1. O lies on the equatorial line of each magnet at distance 10 cm. Equatorial field \( B=\dfrac{\mu_0}{4\pi}\dfrac{M}{r^3} \).
2. Magnet 1: \( 10^{-7}\times1.2/10^{-3}=1.2\times10^{-4} \). Magnet 2: \( 1.0\times10^{-4} \). Both point opposite to the moments, i.e. northward, as does the earth's horizontal field.
3. Total \( (1.2+1.0+0.36)\times10^{-4}=2.56\times10^{-4} \) Wb/m².

Teacher's Note:
1. The equatorial field of a magnet points opposite to its moment.
2. Add all three collinear fields.

Answer: (d) (1) and (3) only

Explanation:
1. The velocity turns towards +y, so the force on the positive charge is initially along +y. \( \hat i\times(-\hat k)=\hat j \), so the field is along -z. (1) is true.
2. The speed is unchanged (4 m/s) and the direction turns by \( \tan^{-1}\dfrac{2}{2\sqrt3}=30^\circ=\dfrac\pi6 \) in 10 ms.
3. Angle \( =\dfrac{QB}{M}t \): \( B=\dfrac{\pi}{6}\cdot\dfrac{M}{Q\times0.01}=\dfrac{50\pi M}{3Q} \). (3) is true.

Teacher's Note:
1. The speed stays the same in a magnetic field; only direction changes.
2. Angle turned equals \( \omega t=\dfrac{QB}{M}t \).

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

Explanation:
1. Inside the hollow cylinder (r < R) the cylinder gives no field, but the solenoid gives \( \mu_0nI \) along the axis, so (1) is true.
2. For R < r < 2R the field is the vector sum of the axial solenoid field and the tangential field \( \dfrac{\mu_0I}{2\pi r} \) of the cylinder, so it is neither purely axial nor purely tangential: (2) and (3) are false.
3. Outside the solenoid its field is zero but the cylinder's current still produces \( \dfrac{\mu_0I}{2\pi r} \), so (4) is true.

Teacher's Note:
1. Apply Ampere's law to each current separately and add the vectors.
2. An ideal solenoid has no field outside, but a straight current always does.

Answer: (a) 3

Explanation:
1. Let \( X_1=x \) so the other wire is at \( 2x \).
2. Same direction: fields oppose between the wires, \( B_1=\dfrac{\mu_0I}{2\pi}\left(\dfrac1x-\dfrac1{2x}\right)=\dfrac{\mu_0I}{4\pi x} \). Opposite directions: fields add, \( B_2=\dfrac{\mu_0I}{2\pi}\left(\dfrac1x+\dfrac1{2x}\right)=\dfrac{3\mu_0I}{4\pi x} \).
3. Radius \( R=\dfrac{mu}{qB}\propto\dfrac1B \), so \( \dfrac{R_1}{R_2}=\dfrac{B_2}{B_1}=3 \).

Teacher's Note:
1. Between parallel currents the fields subtract; between antiparallel currents they add.
2. Radius is inversely proportional to the field.