JEE Physics Moving Charges and Magnetism MCQs Set 19

Multiple Choice Questions (MCQs) for JEE Physics: Moving Charges and Magnetism

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Multiple Choice Questions

Question 1: A charge q moves with speed v in a circle of radius R. The magnitude of the magnetic field it produces at a point on the axis of the circle at distance x from the centre, at the instant considered, is:
(a) \( \frac{\mu_0}{4\pi}\frac{qvR}{(R^2+x^2)^{3/2}} \)
(b) \( \frac{\mu_0}{4\pi}\frac{qv}{R^2+x^2} \)
(c) \( \frac{\mu_0}{4\pi}\frac{qv}{x^2} \)
(d) \( \frac{\mu_0}{4\pi}\frac{qv}{R^2} \)
Show Answer & Explanation

Answer: (b) \( \frac{\mu_0}{4\pi}\frac{qv}{R^2+x^2} \)

Explanation:
1. The distance from the charge to the point is \( r=\sqrt{R^2+x^2} \) and the velocity is perpendicular to the line joining them (velocity is tangential, the line has no tangential part).
2. So \( \sin\theta=1 \) and \( B=\frac{\mu_0}{4\pi}\frac{qv}{R^2+x^2} \).
3. The direction of this field keeps turning as the charge goes round, so the field is constant in magnitude only.

Teacher's Note:
1. Sketch the geometry and decide the direction of each contribution before putting in numbers.
2. Straight-wire formulas depend on the angles the ends subtend at the point.

Question 2: Two bent wires carry 10 A each in the xy-plane, with the origin O midway between R(-0.02 m, 0) and S(0.02 m, 0). The first wire comes along the x-axis from far left to R and then turns and runs along the line x = -0.02 m to far below (negative y). The second wire comes along the x-axis from far right to S and then runs along the line x = 0.02 m to far above (positive y). The magnitude of the magnetic field at O is:
(a) \( 5.0\times10^{-5} \) T, out of the plane
(b) \( 1.0\times10^{-4} \) T, into the plane
(c) \( 1.0\times10^{-4} \) T, out of the plane
(d) zero
Show Answer & Explanation

Answer: (c) \( 1.0\times10^{-4} \) T, out of the plane

Explanation:
1. The straight parts along the x-axis point at O along their own line, so they give no field at O.
2. Each vertical part is a semi-infinite wire starting at the foot of the perpendicular from O, at distance \( d=0.02 \) m: \( B=\frac{\mu_0 i}{4\pi d}=\frac{10^{-7}\times10}{0.02}=5\times10^{-5} \) T.
3. Right-hand rule: the downward current on the left and the upward current on the right both give a field out of the page at O, so they add: \( 1.0\times10^{-4} \) T out of the plane.

Teacher's Note:
1. Sketch the geometry and decide the direction of each contribution before putting in numbers.
2. Straight-wire formulas depend on the angles the ends subtend at the point.

Question 3: A conductor consists of an infinite number of adjacent thin wires, each infinitely long and carrying current i, forming an infinite plane sheet; n is the number of wires per unit length across the sheet. The magnetic field at any point in front of the sheet is:
(a) \( \mu_0 n i \)
(b) \( \frac12\mu_0 n i \)
(c) \( 2\mu_0 n i \)
(d) \( \frac{\mu_0 n i}{4} \)
Show Answer & Explanation

Answer: (b) \( \frac12\mu_0 n i \)

Explanation:
1. Use an Amperian rectangle of length l straddling the sheet, with two sides parallel to the sheet at equal distances on both sides; the field is parallel to the sheet and equal in magnitude on the two sides, opposite in sense.
2. Ampere's law: \( B\,l+B\,l=\mu_0 (n l) i \).
3. So \( B=\frac12\mu_0 n i \), independent of the distance.

Teacher's Note:
1. Sketch the geometry and decide the direction of each contribution before putting in numbers.
2. Straight-wire formulas depend on the angles the ends subtend at the point.

Question 4: A tightly wound long solenoid with n turns per unit length lies with its axis parallel to a large metal sheet carrying a uniform surface current, the current through a strip of width dl being K dl. The magnetic field near the centre of the solenoid is found to be zero. The current in the solenoid is:
(a) \( \frac{K}{n} \)
(b) \( \frac{2K}{n} \)
(c) \( \frac{K}{4n} \)
(d) \( \frac{K}{2n} \)
Show Answer & Explanation

Answer: (d) \( \frac{K}{2n} \)

Explanation:
1. The sheet gives a uniform field \( \frac{\mu_0 K}{2} \) parallel to its plane, perpendicular to the current direction.
2. For the net field to vanish the solenoid field must cancel it: \( \mu_0 n i=\frac{\mu_0 K}{2} \).
3. So \( i=\frac{K}{2n} \).

Teacher's Note:
1. Sketch the geometry and decide the direction of each contribution before putting in numbers.
2. Straight-wire formulas depend on the angles the ends subtend at the point.

Question 5: Two large parallel metal sheets carry equal surface currents, each strip of width dl carrying K dl. The upper sheet carries current out of the plane of the paper and the lower sheet carries current into the plane of the paper. A point Q lies between the sheets. The magnetic field at Q is:
(a) \( \frac{\mu_0 K}{2} \), parallel to the sheets
(b) \( \mu_0 K \), parallel to the sheets
(c) zero
(d) \( 2\mu_0 K \), perpendicular to the sheets
Show Answer & Explanation

Answer: (b) \( \mu_0 K \), parallel to the sheets

Explanation:
1. Each large sheet gives a uniform field \( \frac{\mu_0K}{2} \) parallel to itself, with directions given by the right-hand rule.
2. Between the sheets the two fields point the same way (below the upper sheet and above the lower sheet), so they add: \( \mu_0K \).
3. Outside the pair the fields oppose and cancel.

Teacher's Note:
1. Sketch the geometry and decide the direction of each contribution before putting in numbers.
2. Straight-wire formulas depend on the angles the ends subtend at the point.

Question 6: Two long parallel wires, perpendicular to the plane of the paper, carry currents 2.5 A and I in the same direction (into the paper) at points P and Q. The points P, Q and R lie on a line, with R beyond Q so that PR = 5 m and QR = 2 m. An electron moving along the positive x-direction (from P towards R) at \( 4\times10^5 \) m/s experiences a force of \( 3.2\times10^{-20} \) N at R. The value of I is:
(a) 4 A
(b) 2.5 A
(c) 6 A
(d) 1.5 A
Show Answer & Explanation

Answer: (a) 4 A

Explanation:
1. Force \( F=evB \) gives \( B=\frac{3.2\times10^{-20}}{1.6\times10^{-19}\times4\times10^5}=5\times10^{-7} \) T.
2. Both fields at R have the same direction (same-direction currents, R on the same side of both): \( B=2\times10^{-7}\left(\frac{2.5}{5}+\frac{I}{2}\right) \).
3. Setting equal to \( 5\times10^{-7} \): \( 0.5+\frac I2=2.5 \), so \( I=4 \) A.

Teacher's Note:
1. Take the charge and mass ratio first; every circular-motion result follows from it.
2. Check the direction with the right-hand rule for the sign of the charge.

Question 7: A particle of mass m and positive charge q, moving with speed v, enters a region of uniform magnetic field B (into the page) that fills the half-plane below a straight boundary, reaching the boundary heading down and to the right at 45 degrees to it. The time for which it stays in the field is:
(a) \( \frac{\pi m}{2qB} \)
(b) \( \frac{\pi m}{qB} \)
(c) \( \frac{3\pi m}{2qB} \)
(d) \( \frac{2\pi m}{qB} \)
Show Answer & Explanation

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

Explanation:
1. Period of revolution \( T=\frac{2\pi m}{qB} \).
2. The arc is one quarter of a circle (\( 90^\circ \)), so \( t=\frac T4=\frac{\pi m}{2qB} \).

Teacher's Note:
1. Take the charge and mass ratio first; every circular-motion result follows from it.
2. Check the direction with the right-hand rule for the sign of the charge.

Question 8: A particle of mass m and charge q enters a region of uniform magnetic field B perpendicular to its velocity v, which is itself perpendicular to the straight edge of the region. The width of the region is d, slightly smaller than \( \frac{mv}{qB} \). The angle by which the particle is deviated when it leaves the region is nearly:
(a) \( 45^\circ \)
(b) \( 60^\circ \)
(c) \( 90^\circ \)
(d) \( 180^\circ \)
Show Answer & Explanation

Answer: (c) \( 90^\circ \)

Explanation:
1. Radius of the circle \( r=\frac{mv}{qB} \). On leaving the far edge the deviation \( \theta \) satisfies \( \sin\theta=\frac dr \).
2. With \( d\approx r \) (slightly smaller): \( \sin\theta\approx1 \), so \( \theta\approx90^\circ \).

Teacher's Note:
1. Take the charge and mass ratio first; every circular-motion result follows from it.
2. Check the direction with the right-hand rule for the sign of the charge.

Question 9: A particle of mass m and charge q enters a region of uniform magnetic field B perpendicular to its velocity v, which is perpendicular to the edge of the region. If the width of the region is \( d=\frac{mv}{\sqrt2\,qB} \), the angle through which the particle is deviated on leaving is:
(a) \( 30^\circ \)
(b) \( 60^\circ \)
(c) \( 90^\circ \)
(d) \( 45^\circ \)
Show Answer & Explanation

Answer: (d) \( 45^\circ \)

Explanation:
1. Radius \( r=\frac{mv}{qB} \), so \( \sin\theta=\frac dr=\frac1{\sqrt2} \).
2. Hence \( \theta=45^\circ \).

Teacher's Note:
1. Take the charge and mass ratio first; every circular-motion result follows from it.
2. Check the direction with the right-hand rule for the sign of the charge.

Question 10: A tightly wound long solenoid has n turns per unit length, radius r and carries current i. A particle of charge q and mass m is projected from a point on the axis in a direction perpendicular to the axis. The minimum speed for which it just strikes the solenoid is:
(a) \( \frac{q\mu_0 n i\, r}{m} \)
(b) \( \frac{q\mu_0 n i\, r}{2m} \)
(c) \( \frac{q\mu_0 n i\, r}{4m} \)
(d) \( \frac{2q\mu_0 n i\, r}{m} \)
Show Answer & Explanation

Answer: (b) \( \frac{q\mu_0 n i\, r}{2m} \)

Explanation:
1. Inside the solenoid \( B=\mu_0 n i \) along the axis; the particle moves on a circle that passes through its starting point on the axis.
2. The farthest it can get from the axis is the diameter \( 2\rho \) of that circle, so it just reaches the wall when \( 2\rho=r \).
3. With \( \rho=\frac{mv}{qB} \): \( v=\frac{qBr}{2m}=\frac{q\mu_0 n i r}{2m} \).

Teacher's Note:
1. Take the charge and mass ratio first; every circular-motion result follows from it.
2. Check the direction with the right-hand rule for the sign of the charge.

Question 11: An electron beam passes undeflected through perpendicular electric and magnetic fields, with magnetic induction \( 2\times10^{-3} \) T and electric field \( 3.2\times10^{4} \) V/m. The speed of the electrons is:
(a) \( 1.6\times10^{5} \) m/s
(b) \( 6.4\times10^{7} \) m/s
(c) \( 1.6\times10^{7} \) m/s
(d) \( 3.2\times10^{7} \) m/s
Show Answer & Explanation

Answer: (c) \( 1.6\times10^{7} \) m/s

Explanation:
1. No deflection means \( eE=evB \).
2. So \( v=\frac EB=\frac{3.2\times10^4}{2\times10^{-3}}=1.6\times10^7 \) m/s.

Teacher's Note:
1. Take the charge and mass ratio first; every circular-motion result follows from it.
2. Check the direction with the right-hand rule for the sign of the charge.

Question 12: A current i flows through a cylindrical gold strip of radius r with n free electrons per unit volume (electron charge e). A magnetic field B is applied perpendicular to the current, and free electrons pile up on one side until further accumulation stops. The transverse electric field at that stage has magnitude:
(a) \( \frac{B\,i}{n e} \)
(b) \( \frac{B\,i}{n\pi r^2 e} \)
(c) \( \frac{B\,i\,e}{n\pi r^2} \)
(d) \( \frac{i}{n\pi r^2 e B} \)
Show Answer & Explanation

Answer: (b) \( \frac{B\,i}{n\pi r^2 e} \)

Explanation:
1. Magnetic force on an electron \( e v_d B \) is balanced by the electric force \( eE \).
2. So \( E=v_dB=\frac{B\,i}{n\pi r^2 e} \).

Teacher's Note:
1. Take the charge and mass ratio first; every circular-motion result follows from it.
2. Check the direction with the right-hand rule for the sign of the charge.

Question 13: A proton goes undeflected at \( 10^5 \) m/s in crossed electric and magnetic fields (velocity perpendicular to both). With the electric field switched off, it moves in a circle of radius 2 cm (proton mass \( 1.6\times10^{-27} \) kg). The magnetic and electric fields are:
(a) 0.5 T and 50000 V/m
(b) 0.05 T and 500 V/m
(c) 0.1 T and 10000 V/m
(d) 0.05 T and 5000 V/m
Show Answer & Explanation

Answer: (d) 0.05 T and 5000 V/m

Explanation:
1. Circle: \( B=\frac{mv}{qr}=\frac{1.6\times10^{-27}\times10^5}{1.6\times10^{-19}\times0.02}=0.05 \) T.
2. Undeflected: \( E=vB=10^5\times0.05=5000 \) V/m.

Teacher's Note:
1. Take the charge and mass ratio first; every circular-motion result follows from it.
2. Check the direction with the right-hand rule for the sign of the charge.

Question 14: A particle of mass m and charge q is released from the origin in a region where \( \vec B=B_0\hat j \) and \( \vec E=E_0\hat i \). Its speed as a function of its x-coordinate is:
(a) \( \sqrt{\frac{2qE_0x}{m}} \)
(b) \( \sqrt{\frac{qE_0x}{m}} \)
(c) \( \frac{qE_0x}{mB_0} \)
(d) \( \sqrt{\frac{2qE_0x}{m}}\,\frac{E_0}{B_0} \)
Show Answer & Explanation

Answer: (a) \( \sqrt{\frac{2qE_0x}{m}} \)

Explanation:
1. The magnetic force does no work, so only the electric force changes the kinetic energy.
2. Work-energy theorem: \( \frac12mv^2=qE_0x \).
3. So \( v=\sqrt{\frac{2qE_0x}{m}} \).

Teacher's Note:
1. Take the charge and mass ratio first; every circular-motion result follows from it.
2. Check the direction with the right-hand rule for the sign of the charge.

Question 15: A 10 cm long portion of a straight wire carries a current of 10 A in a magnetic field of 0.1 T making an angle of 37 degrees with the wire (take \( \sin37^\circ=0.6 \)). The magnetic force on the wire is:
(a) 0.08 N
(b) 0.06 N
(c) 0.1 N
(d) 0.6 N
Show Answer & Explanation

Answer: (b) 0.06 N

Explanation:
1. \( F=ilB\sin\theta=10\times0.10\times0.1\times0.6=0.06 \) N.

Teacher's Note:
1. Use F = i l B sin(theta) with l the effective straight length between the ends.
2. A closed loop in a uniform field has zero net force.

Question 16: A wire of length \( \ell \) carries a current i along the y-axis in a magnetic field \( \vec B=B_0(\hat i+\hat j+\hat k) \) T. The magnitude of the magnetic force on the wire is:
(a) \( \sqrt3\, i\ell B_0 \)
(b) \( \sqrt2\, i\ell B_0 \)
(c) \( i\ell B_0 \)
(d) \( 2\, i\ell B_0 \)
Show Answer & Explanation

Answer: (b) \( \sqrt2\, i\ell B_0 \)

Explanation:
1. \( \vec F=i\ell\,\hat j\times B_0(\hat i+\hat j+\hat k)=i\ell B_0(-\hat k+\hat i) \).
2. Magnitude \( i\ell B_0\sqrt{1+1}=\sqrt2\,i\ell B_0 \). (The component of B along the wire does not contribute.)

Teacher's Note:
1. Use F = i l B sin(theta) with l the effective straight length between the ends.
2. A closed loop in a uniform field has zero net force.

Question 17: A wire carrying a current i is kept in the xy-plane along the curve \( y=A\sin\left(\frac{2\pi}{\lambda}x\right) \). A uniform magnetic field B acts in the z-direction. The magnitude of the force on the part of the wire between x = 0 and x = \( \frac\lambda2 \) is:
(a) \( iB\lambda \)
(b) \( \frac{iB\lambda}{4} \)
(c) \( 2\pi iBA \)
(d) \( \frac{iB\lambda}{2} \)
Show Answer & Explanation

Answer: (d) \( \frac{iB\lambda}{2} \)

Explanation:
1. For a curved wire in a uniform field \( \vec F=i\,\vec L\times\vec B \) with \( \vec L \) the vector joining the end points.
2. The end points are (0, 0) and \( (\frac\lambda2,0) \), so \( L=\frac\lambda2 \) and \( F=\frac{iB\lambda}{2} \).

Teacher's Note:
1. Use F = i l B sin(theta) with l the effective straight length between the ends.
2. A closed loop in a uniform field has zero net force.

Question 18: The magnetic field in a region is \( \vec B=B_0\left(1-\frac x\ell\right)\hat k \). A square loop of edge \( \ell \) carrying current i is placed with its edges parallel to the x and y axes in the xy-plane, anywhere in this region. The magnitude of the net magnetic force on the loop is:
(a) \( 2iB_0\ell \)
(b) \( iB_0\ell \)
(c) \( \frac{iB_0\ell}{2} \)
(d) zero
Show Answer & Explanation

Answer: (b) \( iB_0\ell \)

Explanation:
1. The forces on the two edges parallel to the x-axis are equal and opposite, since B is the same along each of them.
2. The two edges parallel to the y-axis at \( x_0 \) and \( x_0+\ell \) have forces \( i\ell B(x_0) \) and \( i\ell B(x_0+\ell) \) in opposite directions.
3. Net \( i\ell\,(B(x_0)-B(x_0+\ell))=i\ell B_0\frac{\ell}{\ell}=iB_0\ell \), independent of position.

Teacher's Note:
1. Use F = i l B sin(theta) with l the effective straight length between the ends.
2. A closed loop in a uniform field has zero net force.

Question 19: Three very long wires AB, CD and EF, with identical resistances, are connected in parallel between wires AE and BF of negligible resistance; neighbouring wires are 2 cm apart (CD is the middle one) and the ammeter in the supply reads 60 A. The magnetic force per unit length on the outer wire AB is:
(a) \( 4\times10^{-3} \) N/m
(b) \( 2\times10^{-3} \) N/m
(c) \( 6\times10^{-3} \) N/m
(d) \( 1.2\times10^{-2} \) N/m
Show Answer & Explanation

Answer: (c) \( 6\times10^{-3} \) N/m

Explanation:
1. Identical resistances share the current equally: 20 A in each wire.
2. Attraction by CD (2 cm): \( \frac{\mu_0 i^2}{2\pi(0.02)}=\frac{2\times10^{-7}\times400}{0.02}=4\times10^{-3} \) N/m; by EF (4 cm): \( 2\times10^{-3} \) N/m.
3. Both pull the same way, so the total is \( 6\times10^{-3} \) N/m towards the other wires.

Teacher's Note:
1. Use F = i l B sin(theta) with l the effective straight length between the ends.
2. A closed loop in a uniform field has zero net force.

Question 20: Two long parallel conductors 5 cm apart carry currents \( i_1=10 \) A and \( i_2=20 \) A in the same direction. The work per unit length to be done against the magnetic force to increase their separation to 10 cm is:
(a) \( 2.8\times10^{-5} \) J/m
(b) \( 1.4\times10^{-5} \) J/m
(c) \( 5.5\times10^{-5} \) J/m
(d) \( 4.0\times10^{-4} \) J/m
Show Answer & Explanation

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

Explanation:
1. Force per unit length \( f=\frac{\mu_0 i_1i_2}{2\pi r} \) (attraction).
2. Work \( =\int_{r_1}^{r_2}f\,dr=\frac{\mu_0i_1i_2}{2\pi}\ln\frac{r_2}{r_1}=2\times10^{-7}\times200\times\ln2 \).
3. This is \( 2.77\times10^{-5} \) J/m.

Teacher's Note:
1. Use F = i l B sin(theta) with l the effective straight length between the ends.
2. A closed loop in a uniform field has zero net force.

Question 21: A straight rod of mass m and length \( \ell \) slides on two parallel horizontal plastic rails a distance d apart; the coefficient of friction between rod and rails is \( \mu \) and the rod carries current i. The minimum vertical magnetic field needed to make the rod slide is:
(a) \( \frac{mg}{\mu id} \)
(b) \( \frac{\mu mg}{id} \)
(c) \( \frac{\mu mg}{i\ell} \)
(d) \( \frac{\mu mg}{2id} \)
Show Answer & Explanation

Answer: (b) \( \frac{\mu mg}{id} \)

Explanation:
1. Only the length between the rails, d, carries current in the field, and it is perpendicular to a vertical field: force \( Bid \).
2. To start sliding this force must equal the limiting friction \( \mu mg \).
3. So \( B=\frac{\mu mg}{id} \).

Teacher's Note:
1. Use F = i l B sin(theta) with l the effective straight length between the ends.
2. A closed loop in a uniform field has zero net force.

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