JEE Physics Moving Charges and Magnetism MCQs Set 17

Answer: (c) 2

Explanation:
1. Radius of the arc: \( R=\frac{mv}{eB} \). An electron entering radially leaves radially; the deflection angle \( \delta \) satisfies \( \tan\frac\delta2=\frac rR=\frac{eBr}{mv} \).
2. Time = angle / angular frequency: \( t=\frac{\delta}{eB/m}=\frac{m\delta}{eB} \).
3. With \( \delta=2\tan^{-1}\frac{eBr}{mv} \): \( t=\frac{2m\tan^{-1}(eBr/mv)}{eB} \), so N = 2.

Teacher's Note:
1. The geometry of two intersecting circles gives the half-angle relation.
2. Check the limit of a very strong field: \( \delta\to\pi \).

Answer: (d) 5

Explanation:
1. Equilibrium: weight per unit length equals the repulsive force per unit length, \( \lambda g=\frac{\mu_0i_1i_2}{2\pi d} \).
2. For a small downward displacement x: extra force per length \( \approx\frac{\mu_0i_1i_2}{2\pi d^2}x=\lambda g\frac{x}{d} \), so \( \omega^2=\frac gd=\frac{10}{0.01}=1000 \).
3. Period: \( T^2=\frac{4\pi^2}{\omega^2}=\frac{40}{1000}=0.04 \), so \( T=0.2=\frac15 \) s and N = 5.

Teacher's Note:
1. Currents in opposite directions repel, which supports the upper wire.
2. The restoring force is proportional to displacement for small x.

Answer: (b) 2

Explanation:
1. The strip is a set of thin wires of current \( J_0\,dx \); for a point at height h above its centre, a strip element at x gives a field component parallel to the strip's plane \( \frac{\mu_0J_0\,dx}{2\pi}\frac{h}{h^2+x^2} \).
2. Integrating over the width from \( -\frac h2 \) to \( \frac h2 \): \( B=\frac{\mu_0J_0}{2\pi}\cdot2\tan^{-1}\frac{h/2}{h}=\frac{\mu_0J_0}{\pi}\tan^{-1}\frac12 \).
3. Force per length \( iB \) gives the stated form with N = 2.

Teacher's Note:
1. For an infinitely wide sheet the arctan tends to \( \frac\pi2 \), giving \( \frac{\mu_0J_0}{2} \).
2. Half-width over height gives the argument.

Answer: (c) 4

Explanation:
1. Let the long wire be perpendicular to the plane of the figure and CD lie in the plane, starting at C (distance a from the wire) and running perpendicular to the line from the wire to C. At distance x along CD the distance from the wire is \( \sqrt{a^2+x^2} \) and the component of the field perpendicular to CD is \( \frac{\mu_0i_1}{2\pi}\frac{x}{a^2+x^2} \).
2. Force: \( F=\int_0^\ell i_2\frac{\mu_0i_1}{2\pi}\frac{x\,dx}{a^2+x^2}=\frac{\mu_0i_1i_2}{4\pi}\ln\frac{a^2+\ell^2}{a^2} \).
3. So N = 4.

Teacher's Note:
1. The component of the field along CD gives no force.
2. Substitute \( u=a^2+x^2 \) to integrate.

Answer: (a) \( \frac{q\omega}{2\pi} \) and \( \frac{q\omega r^2}{2} \)

Explanation:
1. Charge passes a point once per revolution, so \( i=\frac{q}{T}=\frac{q\omega}{2\pi} \).
2. Moment: \( \mu=iA=\frac{q\omega}{2\pi}\pi r^2=\frac{q\omega r^2}{2} \).
3. Hence the pair above.

Teacher's Note:
1. The moment can also be found as \( \frac{q}{2m}L \) with \( L=mr^2\omega \).
2. This gives the gyromagnetic ratio q/2m.

Answer: (b) 7.0 \( \mu T \)

Explanation:
1. Turns per unit radial width: \( \frac{N}{b-a} \). Field at the centre: \( B=\frac{\mu_0NI}{2(b-a)}\ln\frac ba \).
2. Substitute: \( \frac{4\pi\times10^{-7}\times100\times8\times10^{-3}}{2\times0.05}\ln2 \).
3. This gives \( B\approx6.97\times10^{-6} \) T, i.e. about 7.0 \( \mu T \).

Teacher's Note:
1. The log comes from summing \( \frac{dr}{r} \) over the winding.
2. Convert mm to metres before substituting.

Answer: (c) 14.7 mA m\( ^2 \)

Explanation:
1. Number of turns between radius r and r + dr: \( \frac{N}{b-a}dr \), each with moment \( I\pi r^2 \).
2. Sum: \( \mu=\frac{N I\pi}{b-a}\int_a^br^2dr=\frac{\pi NI}{3}(a^2+ab+b^2) \).
3. Numerically \( \mu=\frac{\pi\times100\times8\times10^{-3}}{3}(0.0025+0.005+0.01)=1.47\times10^{-2} \) A m\( ^2 \) = 14.7 mA m\( ^2 \).

Teacher's Note:
1. Check by comparing with N loops of average radius: roughly 100 loops of radius 7.5 cm give 14 mA m^2.
2. Ensure the radii are in metres.

Answer: (a) 3

Explanation:
1. Take \( x_1=x_0/3 \): distances from the wires are \( x_0/3 \) and \( 2x_0/3 \).
2. Same direction: fields oppose, \( B_1\propto\frac{1}{x_0/3}-\frac{1}{2x_0/3}=\frac{3-1.5}{x_0}=\frac{1.5}{x_0} \). Opposite directions: fields add, \( B_2\propto\frac{3+1.5}{x_0}=\frac{4.5}{x_0} \).
3. Radius is inversely proportional to B: \( \frac{R_1}{R_2}=\frac{B_2}{B_1}=3 \).

Teacher's Note:
1. Between two wires, same-direction currents give cancelling fields.
2. The speed u is the same in both cases.

Answer: (b) (i) and (ii) only

Explanation:
1. In a non-uniform field the two poles feel different forces, so a net force is generally present.
2. The torque \( \vec\mu\times\vec B \) is present unless the needle is aligned along the field, so both the cases of force with torque and of force without torque (aligned needle) occur.
3. A torque without any force or no force and no torque cannot be guaranteed in a non-uniform field, so only (i) and (ii) are correct.

Teacher's Note:
1. Force on a dipole depends on the gradient of the field; torque on its orientation.
2. In a uniform field there would be torque but no force.

Answer: (d) (ii), (iii) and (iv) only

Explanation:
1. For a loop inside the tube no current is enclosed, so the field is zero everywhere inside: it is constant (equal to zero), so (ii) is correct and (i) is false.
2. In particular it is zero at the axis, so (iii) is correct.
3. Outside, \( B=\frac{\mu_0I}{2\pi r} \) is not zero, so (iv) is correct.

Teacher's Note:
1. Inside a hollow current-carrying cylinder the field vanishes by Ampere's law.
2. A solid conductor would have a linearly increasing field instead.

Answer: (a) (ii) and (iii) only

Explanation:
1. Outside the cable the enclosed net current is zero, so the field is zero: (i) is false.
2. Inside the inner conductor the enclosed current grows with r so B is non-zero except on the axis: (ii) is correct.
3. Between the conductors the enclosed current is the full inner current, and inside the outer conductor it is partly cancelled but not zero: (iii) is correct. (iv) is false.

Teacher's Note:
1. Apply Ampere's law to loops at different radii.
2. Equal and opposite currents give a perfect cancellation outside, a useful property of a coaxial cable.

Answer: (b) (i) and (iii) only

Explanation:
1. For the same kinetic energy, \( r=\frac{\sqrt{2mK}}{qB} \), so deflection (inversely related to radius in a thin region) depends on \( \frac{\sqrt m}{q} \).
2. H: \( \frac{\sqrt1}{1}=1 \); He: \( \frac{\sqrt4}{1}=2 \); O: \( \frac{\sqrt{16}}{2}=2 \).
3. H has the smallest radius, so it is deflected most; He and O have equal radii and are deflected equally: (i) and (iii).

Teacher's Note:
1. A smaller radius means a larger deflection angle over a fixed thickness.
2. Compare sqrt(m)/q.

Answer: (d) (ii) and (iv) only

Explanation:
1. Radius \( r=\frac{mv}{qB} \) is inversely proportional to the charge: the singly ionised ion has radius three times that of the triply ionised one, so (i) is false and (ii) is true.
2. Both start at the same point with the same velocity and curve the same way, so their circles have a common tangent there and touch each other: (iv) is true and (iii) is false.

Teacher's Note:
1. Draw two circles of different radii tangent at the starting point, on the same side.
2. Same sign of charge is assumed for the two ions.

Answer: (a) (iii) and (iv) only

Explanation:
1. (i) possible: the magnetic force is zero and the electric force is along the velocity, so the direction is unchanged.
2. (ii) possible: with \( \vec E\perp\vec B \) and \( v=E/B \) the forces cancel.
3. (iii) not possible: the magnetic force is zero but the electric force is sideways, so the particle deflects. (iv) not possible: a perpendicular velocity component gives a sideways magnetic force that cannot be cancelled by an electric force along B.

Teacher's Note:
1. The question asks for the impossible cases.
2. Resolve the velocity along and perpendicular to B.

Answer: (c) (i) and (iii) only

Explanation:
1. Each sheet gives \( \frac{\mu_0K}{2}=\frac{4\pi\times10^{-7}}{2}\cdot\frac{10}{\pi}=2\,\mu T \) on either side.
2. Above both sheets the fields add: \( 4\,\mu T \), and for currents into the page the field above is towards +x: (i) is correct.
3. Between the sheets the fields point in opposite directions and cancel, so the field at B is zero: (iii) is correct.

Teacher's Note:
1. Treat each sheet as an infinite plane of current with uniform field on each side.
2. The field direction follows the right-hand rule for a current into the page.

Answer: (d) (ii) and (iii) only

Explanation:
1. From \( F=\frac{q_1q_2}{4\pi\varepsilon_0r^2} \): \( [\varepsilon_0]=\frac{(IT)^2}{MLT^{-2}L^2}=M^{-1}L^{-3}T^4I^2 \), so (ii) is correct.
2. From \( \frac FL=\frac{\mu_0i^2}{2\pi d} \): \( [\mu_0]=\frac{MLT^{-2}}{I^2} \), so (iii) is correct.
3. (i) and (iv) do not match these results.

Teacher's Note:
1. Check with \( c^2=\frac{1}{\mu_0\varepsilon_0} \): the product has dimensions \( L^{-2}T^2 \).
2. Charge has dimension IT.

Answer: (a) (i) and (ii) only

Explanation:
1. The magnetic force is perpendicular to both v and B, so B must be perpendicular to x.
2. A field along either y or z turns the velocity within a plane perpendicular to B; applied for half a revolution, it reverses the velocity.
3. So both (i) and (ii) work; the 'only' statements are wrong.

Teacher's Note:
1. Any direction perpendicular to the velocity works, with the sense of turning decided by B.
2. A field along x would not change the velocity.

Answer: (c) (ii) and (iv) only

Explanation:
1. Each particle makes a semicircle and leaves the region, so (i) is false.
2. Both leave with velocity reversed (opposite to the entry velocity), so they emerge along parallel paths: (ii) is true.
3. The time for a semicircle is \( \frac{\pi m}{qB} \), which differs for the electron and proton because the masses differ: (iii) is false and (iv) is true.

Teacher's Note:
1. The charges have equal magnitude, so the time depends on mass only.
2. Their circles bend in opposite directions.

Answer: (d) (iii) and (iv) only

Explanation:
1. For \( \theta=0^\circ \) the velocity is perpendicular to B, giving circular motion in the x-z plane, but E accelerates the charge along y, so the path is a helix with increasing pitch: (i) and (ii) are false.
2. For \( \theta=10^\circ \) the velocity has components perpendicular and parallel to B; the motion is a helix about the y-direction whose pitch grows as the y-velocity increases under E: (iii) is true.
3. For \( \theta=90^\circ \) the velocity is along B, so the magnetic force is zero and the charge moves along y with constant acceleration: (iv) is true.

Teacher's Note:
1. E and B are parallel, so E acts only along the helix axis.
2. Pitch is proportional to the axial speed.

Answer: (b) (i) and (iv) only

Explanation:
1. Inside the solenoid there is an axial field \( \mu_0nI \) everywhere (r < 2R); the hollow cylinder adds nothing inside it. So for 0 < r < R the field is non-zero: (i) is correct.
2. For R < r < 2R both the axial solenoid field and the tangential field of the cylinder are present, so the field is neither purely axial nor purely tangential: (ii) and (iii) are false.
3. For r > 2R the solenoid field vanishes, but the cylinder gives \( \frac{\mu_0I}{2\pi r} \) (tangential), so the field is non-zero: (iv) is correct.

Teacher's Note:
1. Superpose the two fields.
2. An ideal solenoid has no field outside.

Answer: (c) (ii) and (iii) only

Explanation:
1. Drift speed \( v_d=\frac{I}{newd} \) and the Hall field \( E=v_dB \), so \( V=Ed=v_dBd=\frac{BI}{new} \).
2. V depends on the width but not on the thickness. If the widths are equal, \( V_2=V_1 \): (ii) is correct.
3. If \( w_1=2w_2 \), \( V_2=\frac{BI}{new_2}=2V_1 \): (iii) is correct.

Teacher's Note:
1. The thickness d cancels out of the Hall voltage.
2. V is inversely proportional to the width.

Answer: (d) (i) and (iii) only

Explanation:
1. Hall voltage \( V=\frac{BI}{new} \), so \( V\propto\frac Bn \) for the same I, e and w.
2. With equal B and \( n_1=2n_2 \): \( V_2=2V_1 \): (i) is correct.
3. With \( B_1=2B_2 \) and equal n: \( V_2=\frac{V_1}{2} \): (iii) is correct.

Teacher's Note:
1. Write V as proportional to B/n and compare ratios.
2. Larger carrier density means smaller Hall voltage.

Answer: (b) Electron with \( \vec v=\frac{E_0}{B_0}\hat y \), \( \vec E=-E_0\hat x \), \( \vec B=B_0\hat z \)

Explanation:
1. Electron: electric force \( -e\vec E=+eE_0\hat x \); magnetic force \( -e\vec v\times\vec B=-e\frac{E_0}{B_0}B_0(\hat y\times\hat z)=-eE_0\hat x \). They cancel, so the velocity is constant.
2. The first proton option has E along z, which accelerates the particle along z, so its velocity changes.
3. The two proton-at-rest options are accelerated by E and so do not keep constant velocity.

Teacher's Note:
1. Constant velocity needs zero net force.
2. This is the velocity selector condition.

Answer: (c) Proton with \( \vec v=2\frac{E_0}{B_0}\hat x \), \( \vec E=E_0\hat z \), \( \vec B=B_0\hat z \)

Explanation:
1. A helix about the z-axis needs B along z so that motion perpendicular to z is circular, plus acceleration or drift along +z.
2. In the first option B is along z, the velocity is along x (perpendicular to B) and E is along +z: a circle in the xy-plane combined with acceleration along +z, giving a helix about +z.
3. The other options have B along y or x, so their helical axes are along y or x.

Teacher's Note:
1. The helix axis is along B when E is parallel to B.
2. A positive charge accelerates along E.