JEE Physics Moving Charges and Magnetism MCQs Set 31

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Chapter-wise Objective Questions: Moving Charges and Magnetism

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Multiple Choice Questions

Question 1: One of two identical conducting wires of length L is bent into a circular loop and the other into a circular coil of N identical turns. The same current is passed in both. The ratio of the magnetic field at the centre of the loop to that at the centre of the coil, \( B_L/B_C \), is
(a) \( \dfrac1{N^2} \)
(b) \( \dfrac1N \)
(c) \( N \)
(d) \( N^2 \)
Show Answer & Explanation

Answer: (a) \( \dfrac1{N^2} \)

Explanation:
1. Loop: \( L=2\pi r \), \( B_L=\dfrac{\mu_0I}{2r} \). Coil: \( L=N\cdot2\pi r' \Rightarrow r'=\dfrac rN \).
2. \( B_C=N\dfrac{\mu_0I}{2r'}=\dfrac{N^2\mu_0I}{2r} \), so \( B_L/B_C=1/N^2 \).

Teacher's Note:
1. Write the Biot-Savart or Ampere result for each element of the geometry and add the fields as vectors.
2. Check that the answer reduces correctly in a limiting case such as zero current or a straight infinite wire.

Question 2: A current loop has two circular arcs of radii 3 cm and 5 cm with a common centre O, each subtending \( 45^\circ \) at O, joined by two radial straight lines. The loop carries 10 A, flowing in opposite senses in the two arcs. The magnetic field at O is close to (\( \mu_0=4\pi\times10^{-7} \))
(a) \( 1.0\times10^{-7}\ \text T \)
(b) \( 1.0\times10^{-5}\ \text T \)
(c) \( 1.5\times10^{-5}\ \text T \)
(d) \( 1.5\times10^{-7}\ \text T \)
Show Answer & Explanation

Answer: (b) \( 1.0\times10^{-5}\ \text T \)

Explanation:
1. The radial segments contribute nothing at O. Each arc gives \( \dfrac{\mu_0I}{4\pi}\dfrac{\theta}{r} \) with \( \theta=\pi/4 \).
2. Opposite senses subtract: \( B=10^{-7}\times10\times\dfrac\pi4\left(\dfrac1{0.03}-\dfrac1{0.05}\right)=1.0\times10^{-5} \) T.

Teacher's Note:
1. Write the Biot-Savart or Ampere result for each element of the geometry and add the fields as vectors.
2. Check that the answer reduces correctly in a limiting case such as zero current or a straight infinite wire.

Question 3: A long insulated copper wire is closely wound as a flat spiral of N turns, with inner radius a and outer radius b, lying in the xy plane and carrying a steady current I. The z-component of the magnetic field at the centre of the spiral is
(a) \( \dfrac{\mu_0NI}{2(b-a)}\ln\dfrac{b+a}{b-a} \)
(b) \( \dfrac{\mu_0NI}{2b}\ln\dfrac ba \)
(c) \( \dfrac{\mu_0NI}{2b}\ln\dfrac{b+a}{b-a} \)
(d) \( \dfrac{\mu_0NI}{2(b-a)}\ln\dfrac ba \)
Show Answer & Explanation

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

Explanation:
1. Turn density: \( \dfrac{N}{b-a} \) per unit radius, so a ring between r and r + dr has \( dN=\dfrac{N\,dr}{b-a} \) turns.
2. \( B=\int_a^b\dfrac{\mu_0I\,dN}{2r}=\dfrac{\mu_0NI}{2(b-a)}\int_a^b\dfrac{dr}{r}=\dfrac{\mu_0NI}{2(b-a)}\ln\dfrac ba \).

Teacher's Note:
1. Write the Biot-Savart or Ampere result for each element of the geometry and add the fields as vectors.
2. Check that the answer reduces correctly in a limiting case such as zero current or a straight infinite wire.

Question 4: A long straight wire along the z-axis carries a current I in the negative z-direction. The magnetic field \( \vec B \) at a point (x, y) in the z = 0 plane is
(a) \( \dfrac{\mu_0I(y\hat i-x\hat j)}{2\pi(x^2+y^2)} \)
(b) \( \dfrac{\mu_0I(x\hat i+y\hat j)}{2\pi(x^2+y^2)} \)
(c) \( \dfrac{\mu_0I(x\hat j-y\hat i)}{2\pi(x^2+y^2)} \)
(d) \( \dfrac{\mu_0I(x\hat i-y\hat j)}{2\pi(x^2+y^2)} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{\mu_0I(y\hat i-x\hat j)}{2\pi(x^2+y^2)} \)

Explanation:
1. For a current along \( +z \) the field circulates anticlockwise: direction \( \dfrac{-y\hat i+x\hat j}{\sqrt{x^2+y^2}} \).
2. Reversing the current reverses the field: \( \vec B=\dfrac{\mu_0I}{2\pi(x^2+y^2)}(y\hat i-x\hat j) \).

Teacher's Note:
1. Write the Biot-Savart or Ampere result for each element of the geometry and add the fields as vectors.
2. Check that the answer reduces correctly in a limiting case such as zero current or a straight infinite wire.

Question 5: A coil of N turns is wound tightly in the form of a spiral with inner and outer radii a and b. When a current I passes through it, the magnetic field at the centre is
(a) \( \dfrac{\mu_0NI}{b} \)
(b) \( \dfrac{\mu_0NI}{2(b-a)}\ln\dfrac ba \)
(c) \( \dfrac{2\mu_0NI}{a} \)
(d) \( \dfrac{\mu_0I^N}{2(b-a)}\ln\dfrac ba \)
Show Answer & Explanation

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

Explanation:
1. Summing rings of radius r with turn density \( N/(b-a) \): \( B=\int\dfrac{\mu_0I}{2r}\dfrac{N\,dr}{b-a} \).
2. This gives \( \dfrac{\mu_0NI}{2(b-a)}\ln\dfrac ba \).

Teacher's Note:
1. Write the Biot-Savart or Ampere result for each element of the geometry and add the fields as vectors.
2. Check that the answer reduces correctly in a limiting case such as zero current or a straight infinite wire.

Question 6: A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB subtends an angle \( \theta \) at the centre. The magnetic induction at the centre due to the current in the ring is
(a) proportional to \( (180^\circ-\theta) \)
(b) inversely proportional to r
(c) zero only if \( \theta=180^\circ \)
(d) zero for all values of \( \theta \)
Show Answer & Explanation

Answer: (d) zero for all values of \( \theta \)

Explanation:
1. The two arcs are in parallel; the currents satisfy \( I_1\theta=I_2(2\pi-\theta) \) (equal potential drop, resistance proportional to length).
2. Field of an arc \( =\dfrac{\mu_0I\theta}{4\pi r} \). The two arcs give equal and opposite fields \( \dfrac{\mu_0I_1\theta}{4\pi r}-\dfrac{\mu_0I_2(2\pi-\theta)}{4\pi r}=0 \) for every \( \theta \).

Teacher's Note:
1. Write the Biot-Savart or Ampere result for each element of the geometry and add the fields as vectors.
2. Check that the answer reduces correctly in a limiting case such as zero current or a straight infinite wire.

Question 7: A current I flows along the length of an infinitely long, straight, thin-walled pipe. Then the magnetic field
(a) at any point inside the pipe is zero
(b) at all points inside the pipe is the same but not zero
(c) is zero only on the axis of the pipe
(d) is different at different points inside the pipe
Show Answer & Explanation

Answer: (a) at any point inside the pipe is zero

Explanation:
1. By Ampere's law, a circular loop inside the pipe encloses no current.
2. By symmetry B is circumferential and constant on the loop, hence \( B\cdot2\pi r=0 \Rightarrow B=0 \) everywhere inside.

Teacher's Note:
1. Write the Biot-Savart or Ampere result for each element of the geometry and add the fields as vectors.
2. Check that the answer reduces correctly in a limiting case such as zero current or a straight infinite wire.

Question 8: Two parallel wires in the plane of the paper are a distance \( x_0 \) apart. A point charge moves with speed u between them, in the same plane, at distance \( x_1 \) from one wire. When the wires carry current I in the same direction the radius of curvature of the path is \( R_1 \); when the currents are in opposite directions it is \( R_2 \). If \( x_0/x_1=3 \), then \( R_1/R_2 \) is
(a) 1/3
(b) 2
(c) 3
(d) 3/2
Show Answer & Explanation

Answer: (c) 3

Explanation:
1. \( R=\dfrac{mu}{qB}\propto\dfrac1B \). With \( x_0=3x_1 \) the distances from the wires are \( x_1 \) and \( 2x_1 \).
2. Same direction: \( B_1\propto\dfrac1{x_1}-\dfrac1{2x_1}=\dfrac1{2x_1} \). Opposite: \( B_2\propto\dfrac1{x_1}+\dfrac1{2x_1}=\dfrac3{2x_1} \). \( \dfrac{R_1}{R_2}=\dfrac{B_2}{B_1}=3 \).

Teacher's Note:
1. Write the Biot-Savart or Ampere result for each element of the geometry and add the fields as vectors.
2. Check that the answer reduces correctly in a limiting case such as zero current or a straight infinite wire.

Question 9: Two long parallel wires A and B, a distance d apart, carry currents \( I_1 \) (upwards) and \( I_2 \) (downwards). A third wire C carrying a current I is placed parallel to them at a distance x from A so that the net force on it is zero. The possible values of x are
(a) \( x=\dfrac{I_1}{I_1+I_2}d\ \text{and}\ x=\dfrac{I_2}{I_1-I_2}d \)
(b) \( x=\dfrac{I_2}{I_1+I_2}d\ \text{and}\ x=\dfrac{I_2}{I_1-I_2}d \)
(c) \( x=\dfrac{I_1}{I_1-I_2}d\ \text{only} \)
(d) \( x=\pm\dfrac{I_1}{I_1-I_2}d \)
Show Answer & Explanation

Answer: (d) \( x=\pm\dfrac{I_1}{I_1-I_2}d \)

Explanation:
1. Antiparallel currents: the fields of A and B cancel only outside the wires, on the side of the smaller current.
2. For \( I_1\gt I_2 \), beyond B: \( \dfrac{I_1}{x}=\dfrac{I_2}{x-d}\Rightarrow x=\dfrac{I_1d}{I_1-I_2} \). For \( I_1\lt I_2 \) the point lies on the other side of A at \( x=-\dfrac{I_1d}{I_2-I_1} \), so \( x=\pm\dfrac{I_1d}{I_1-I_2} \).

Teacher's Note:
1. Use \( F=BIl\sin\theta \) or \( F/l=\mu_0I_1I_2/2\pi d \) and decide attraction or repulsion from current directions.
2. Check the result with a symmetric or limiting case.

Question 10: A circular coil of N turns and radius r carries a current I. It is held in the XZ plane in a magnetic field \( B\hat i \). The torque on the coil, in N m, is
(a) \( B\pi r^2IN \)
(b) \( \dfrac{Br^2I}{\pi N} \)
(c) \( \dfrac{B\pi r^2I}{N} \)
(d) zero
Show Answer & Explanation

Answer: (a) \( B\pi r^2IN \)

Explanation:
1. The coil's moment is along y (normal to the XZ plane), perpendicular to B along x.
2. \( \tau=NIAB\sin90^\circ=N I\pi r^2B \).

Teacher's Note:
1. Use \( F=BIl\sin\theta \) or \( F/l=\mu_0I_1I_2/2\pi d \) and decide attraction or repulsion from current directions.
2. Check the result with a symmetric or limiting case.

Question 11: Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium, each thread making an angle \( \theta \) with the vertical. If the wires have mass \( \lambda \) per unit length, then I is (g is the acceleration due to gravity)
(a) \( \sin\theta\sqrt{\dfrac{\pi\lambda gL}{\mu_0\cos\theta}} \)
(b) \( 2\sqrt{\dfrac{\pi gL}{\mu_0}\tan\theta} \)
(c) \( 2\sin\theta\sqrt{\dfrac{\pi\lambda gL}{\mu_0\cos\theta}} \)
(d) \( \sqrt{\dfrac{\pi\lambda gL}{\mu_0}\tan\theta} \)
Show Answer & Explanation

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

Explanation:
1. Wires repel; separation \( d=2L\sin\theta \). Per unit length: \( \dfrac{\mu_0I^2}{2\pi d}=\lambda g\tan\theta \).
2. \( I^2=\dfrac{2\pi d\lambda g\tan\theta}{\mu_0}=\dfrac{4\pi\lambda gL\sin\theta\tan\theta}{\mu_0} \), so \( I=2\sin\theta\sqrt{\dfrac{\pi\lambda gL}{\mu_0\cos\theta}} \).

Teacher's Note:
1. Use \( F=BIl\sin\theta \) or \( F/l=\mu_0I_1I_2/2\pi d \) and decide attraction or repulsion from current directions.
2. Check the result with a symmetric or limiting case.

Question 12: Two thin long parallel wires separated by a distance b carry a current i ampere each. The magnitude of the force per unit length exerted by one wire on the other is
(a) \( \dfrac{\mu_0i^2}{b^2} \)
(b) \( \dfrac{\mu_0i}{2\pi b} \)
(c) \( \dfrac{\mu_0i}{2\pi b^2} \)
(d) \( \dfrac{\mu_0i^2}{2\pi b} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{\mu_0i^2}{2\pi b} \)

Explanation:
1. Field of one wire at the other: \( B=\dfrac{\mu_0i}{2\pi b} \).
2. Force per unit length \( =iB=\dfrac{\mu_0i^2}{2\pi b} \).

Teacher's Note:
1. Use \( F=BIl\sin\theta \) or \( F/l=\mu_0I_1I_2/2\pi d \) and decide attraction or repulsion from current directions.
2. Check the result with a symmetric or limiting case.

Question 13: Is it true that no net force acts on a rectangular coil carrying a steady current when it is suspended freely in a uniform magnetic field?
(a) No, a net force always acts along the normal to the coil
(b) Yes, the forces on opposite sides are equal and opposite, so only a torque can act
(c) No, the net force is along the field direction
(d) No, a net force acts unless the field is parallel to the plane
Show Answer & Explanation

Answer: (b) Yes, the forces on opposite sides are equal and opposite, so only a torque can act

Explanation:
1. In a uniform field the force on each side is \( I\vec l\times\vec B \); the vector sum of the four sides is \( I(\sum\vec l)\times\vec B=0 \).
2. The forces form a couple, so the coil may rotate but its centre of mass does not accelerate.

Teacher's Note:
1. Use \( F=BIl\sin\theta \) or \( F/l=\mu_0I_1I_2/2\pi d \) and decide attraction or repulsion from current directions.
2. Check the result with a symmetric or limiting case.

Question 14: A long horizontal wire AB, free to move in a vertical plane and carrying a steady current of 20 A, is in equilibrium at a height of 0.01 m over another parallel long wire CD which is fixed in a horizontal plane and carries a steady current of 30 A. If AB is slightly depressed it executes simple harmonic motion. The period of oscillation is close to (g = 9.8 m/s\(^2\))
(a) 0.02 s
(b) 2 s
(c) 0.2 s
(d) 0.4 s
Show Answer & Explanation

Answer: (c) 0.2 s

Explanation:
1. Equilibrium: repulsion per length \( \dfrac{\mu_0I_1I_2}{2\pi h}=\lambda g \).
2. For a small displacement x the restoring force per length is \( \dfrac{\mu_0I_1I_2}{2\pi h^2}x=\dfrac{\lambda g}{h}x \), so \( \omega^2=\dfrac gh \) and \( T=2\pi\sqrt{h/g}=2\pi\sqrt{0.01/9.8}=0.2 \) s.

Teacher's Note:
1. Use \( F=BIl\sin\theta \) or \( F/l=\mu_0I_1I_2/2\pi d \) and decide attraction or repulsion from current directions.
2. Check the result with a symmetric or limiting case.

Question 15: Three infinitely long thin wires, each carrying current i in the same direction, lie in the xy plane in a gravity-free space. The central wire is along the y-axis and the other two along \( x=\pm d \). If the central wire (linear density \( \lambda \)) is displaced slightly along the z-direction and released, it executes simple harmonic motion with frequency
(a) \( \dfrac{i}{\pi d}\sqrt{\dfrac{\mu_0}{\pi\lambda}} \)
(b) \( \dfrac{i}{2\pi d}\sqrt{\dfrac{\mu_0}{2\pi\lambda}} \)
(c) \( \dfrac{i}{4\pi d}\sqrt{\dfrac{\mu_0}{\pi\lambda}} \)
(d) \( \dfrac{i}{2\pi d}\sqrt{\dfrac{\mu_0}{\pi\lambda}} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{i}{2\pi d}\sqrt{\dfrac{\mu_0}{\pi\lambda}} \)

Explanation:
1. For displacement z the distance to each outer wire is \( \sqrt{d^2+z^2} \); each attracts with \( \dfrac{\mu_0i^2}{2\pi\sqrt{d^2+z^2}} \) per length, and the z-component is \( \dfrac{\mu_0i^2z}{2\pi(d^2+z^2)}\approx\dfrac{\mu_0i^2z}{2\pi d^2} \).
2. Two wires: restoring force per length \( \dfrac{\mu_0i^2}{\pi d^2}z \), so \( \omega^2=\dfrac{\mu_0i^2}{\pi\lambda d^2} \) and \( f=\dfrac{\omega}{2\pi}=\dfrac{i}{2\pi d}\sqrt{\dfrac{\mu_0}{\pi\lambda}} \).

Teacher's Note:
1. Use \( F=BIl\sin\theta \) or \( F/l=\mu_0I_1I_2/2\pi d \) and decide attraction or repulsion from current directions.
2. Check the result with a symmetric or limiting case.

Question 16: Wire A (9.6 A into the paper) and wire B (3 A out of the paper) are 2 m apart. A point S is 1.6 m from A and 1.2 m from B (so that \( AS\perp BS \)). The magnitude of the magnetic field at S is
(a) \( 7\times10^{-7}\ \text T \)
(b) \( 13\times10^{-7}\ \text T \)
(c) \( 17\times10^{-7}\ \text T \)
(d) \( 22\times10^{-7}\ \text T \)
Show Answer & Explanation

Answer: (b) \( 13\times10^{-7}\ \text T \)

Explanation:
1. \( B_A=\dfrac{\mu_0I_A}{2\pi\,AS}=\dfrac{2\times10^{-7}\times9.6}{1.6}=12\times10^{-7} \) T, \( B_B=\dfrac{2\times10^{-7}\times3}{1.2}=5\times10^{-7} \) T.
2. Each field is perpendicular to the line joining the wire to S, and \( AS\perp BS \), so the fields are perpendicular: \( B=\sqrt{12^2+5^2}\times10^{-7}=13\times10^{-7} \) T.

Teacher's Note:
1. Use \( F=BIl\sin\theta \) or \( F/l=\mu_0I_1I_2/2\pi d \) and decide attraction or repulsion from current directions.
2. Check the result with a symmetric or limiting case.

Question 17: For the same wires (A: 9.6 A into the paper; B: 3 A out of the paper; 2 m apart), the force per unit length on wire B is
(a) \( 5.76\times10^{-6}\ \text{N/m} \)
(b) \( 1.44\times10^{-6}\ \text{N/m} \)
(c) \( 2.88\times10^{-6}\ \text{N/m} \)
(d) \( 9.6\times10^{-7}\ \text{N/m} \)
Show Answer & Explanation

Answer: (c) \( 2.88\times10^{-6}\ \text{N/m} \)

Explanation:
1. \( \dfrac Fl=\dfrac{\mu_0I_AI_B}{2\pi d}=\dfrac{2\times10^{-7}\times9.6\times3}{2} \).
2. \( =2.88\times10^{-6} \) N/m (repulsive, since the currents are opposite).

Teacher's Note:
1. Use \( F=BIl\sin\theta \) or \( F/l=\mu_0I_1I_2/2\pi d \) and decide attraction or repulsion from current directions.
2. Check the result with a symmetric or limiting case.

Question 18: An insulating thin rod of length l has a linear charge density \( \rho(x)=\rho_0\dfrac xl \) on it. The rod is rotated about an axis passing through the origin (x = 0) and perpendicular to the rod, making n rotations per second. The time-averaged magnetic moment of the rod is
(a) \( \dfrac\pi4n\rho_0l^3 \)
(b) \( n\rho_0l^3 \)
(c) \( \pi n\rho_0l^3 \)
(d) \( \dfrac\pi3n\rho_0l^3 \)
Show Answer & Explanation

Answer: (a) \( \dfrac\pi4n\rho_0l^3 \)

Explanation:
1. A charge element \( dq=\rho\,dx \) at distance x forms a current \( n\,dq \) over area \( \pi x^2 \): \( dm=n\pi x^2\rho_0\dfrac xl\,dx \).
2. \( m=\dfrac{n\pi\rho_0}{l}\int_0^lx^3dx=\dfrac{\pi n\rho_0l^3}{4} \).

Teacher's Note:
1. Express the quantity through the magnetic moment \( m=IA \) and the angular momentum \( L \), or use \( \vec\tau=\vec m\times\vec B \).
2. Cross-check by dimensions or a special case.

Question 19: An infinitely long current-carrying wire and a small circular current loop of radius a are in the same plane. The distance of the loop centre from the wire is d (\( d\gg a \)). If the loop applies a force F on the wire, then
(a) \( F\propto\dfrac{a^2}{d^3} \)
(b) \( F\propto\left(\dfrac ad\right)^2 \)
(c) \( F=0 \)
(d) \( F\propto\dfrac ad \)
Show Answer & Explanation

Answer: (b) \( F\propto\left(\dfrac ad\right)^2 \)

Explanation:
1. The loop is a dipole of moment \( m=I\pi a^2 \). In the non-uniform field of the wire, \( F=m\left|\dfrac{dB}{dx}\right| \).
2. With \( B=\dfrac{\mu_0I'}{2\pi x} \), \( \dfrac{dB}{dx}\propto\dfrac1{d^2} \), so \( F\propto\dfrac{a^2}{d^2}=\left(\dfrac ad\right)^2 \).

Teacher's Note:
1. Express the quantity through the magnetic moment \( m=IA \) and the angular momentum \( L \), or use \( \vec\tau=\vec m\times\vec B \).
2. Cross-check by dimensions or a special case.

Question 20: The dipole moment of a circular loop carrying a current I is m, and the magnetic field at the centre of the loop is \( B_1 \). When the dipole moment is doubled by keeping the current constant, the field at the centre is \( B_2 \). The ratio \( B_1/B_2 \) is
(a) \( \dfrac1{\sqrt2} \)
(b) 2
(c) \( \sqrt2 \)
(d) \( \sqrt3 \)
Show Answer & Explanation

Answer: (c) \( \sqrt2 \)

Explanation:
1. \( m=I\pi r^2 \) and \( B=\dfrac{\mu_0I}{2r} \). Doubling m at constant I gives \( r'=\sqrt2\,r \).
2. \( B_2=\dfrac{B_1}{\sqrt2} \), so \( B_1/B_2=\sqrt2 \).

Teacher's Note:
1. Express the quantity through the magnetic moment \( m=IA \) and the angular momentum \( L \), or use \( \vec\tau=\vec m\times\vec B \).
2. Cross-check by dimensions or a special case.

Question 21: Two particles, each of mass m and charge q, are attached to the two ends of a light rigid rod of length 2R. The rod is rotated at constant angular speed about a perpendicular axis through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre is
(a) \( \dfrac q{2m} \)
(b) \( \dfrac qm \)
(c) \( \dfrac{2q}m \)
(d) \( \dfrac q{\pi m} \)
Show Answer & Explanation

Answer: (a) \( \dfrac q{2m} \)

Explanation:
1. Each charge: \( \mu=\dfrac{q\omega R^2}{2} \); two charges: \( \mu=q\omega R^2 \). Angular momentum \( L=2mR^2\omega \).
2. \( \dfrac\mu L=\dfrac{q}{2m} \).

Teacher's Note:
1. Express the quantity through the magnetic moment \( m=IA \) and the angular momentum \( L \), or use \( \vec\tau=\vec m\times\vec B \).
2. Cross-check by dimensions or a special case.

Question 22: A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field \( \vec B_0 \) perpendicular to the plane of the loop. The magnetic force acting on the loop is
(a) \( ir\vec B_0 \)
(b) zero
(c) \( 2\pi ir\vec B_0 \)
(d) \( \pi ir\vec B_0 \)
Show Answer & Explanation

Answer: (b) zero

Explanation:
1. Net force in a uniform field \( =i\left(\oint d\vec l\right)\times\vec B_0 \).
2. \( \oint d\vec l=0 \) for a closed loop, so the net force is zero (each element is stretched radially, giving only tension).

Teacher's Note:
1. Express the quantity through the magnetic moment \( m=IA \) and the angular momentum \( L \), or use \( \vec\tau=\vec m\times\vec B \).
2. Cross-check by dimensions or a special case.

Question 23: A wire of length L metre carrying a current i ampere is bent in the form of a circle. The magnitude of its magnetic moment in MKS units is
(a) \( \dfrac{Li}{2\pi} \)
(b) \( \dfrac{L^2i}{2\pi} \)
(c) \( \dfrac{L^2i}{\pi} \)
(d) \( \dfrac{L^2i}{4\pi} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{L^2i}{4\pi} \)

Explanation:
1. \( 2\pi r=L\Rightarrow r=\dfrac L{2\pi} \).
2. \( m=i\pi r^2=i\pi\dfrac{L^2}{4\pi^2}=\dfrac{L^2i}{4\pi} \).

Teacher's Note:
1. Express the quantity through the magnetic moment \( m=IA \) and the angular momentum \( L \), or use \( \vec\tau=\vec m\times\vec B \).
2. Cross-check by dimensions or a special case.

Question 24: An electron in the ground state of the hydrogen atom revolves anticlockwise in a circular orbit of radius R. Its orbital magnetic moment is (h is Planck's constant, m the mass of the electron)
(a) \( \dfrac{eh}{4\pi m} \)
(b) \( \dfrac{eh}{2\pi m} \)
(c) \( \dfrac{eh}{8\pi m} \)
(d) \( \dfrac{2eh}{\pi m} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{eh}{4\pi m} \)

Explanation:
1. Magnetic moment \( \mu=\dfrac{e}{2m}L \). Bohr quantisation: \( L=\dfrac{nh}{2\pi}=\dfrac h{2\pi} \) for the ground state.
2. \( \mu=\dfrac{e}{2m}\dfrac{h}{2\pi}=\dfrac{eh}{4\pi m} \).

Teacher's Note:
1. Express the quantity through the magnetic moment \( m=IA \) and the angular momentum \( L \), or use \( \vec\tau=\vec m\times\vec B \).
2. Cross-check by dimensions or a special case.

Question 25: The hydrogen atom in its ground state (orbital magnetic moment \( \dfrac{eh}{4\pi m} \)) is placed in a uniform magnetic induction B such that the normal to the plane of the electron's orbit makes \( 30^\circ \) with B. The torque on the orbiting electron is
(a) \( \dfrac{ehB}{4\pi m} \)
(b) \( \dfrac{ehB}{8\pi m} \)
(c) \( \dfrac{\sqrt3\,ehB}{8\pi m} \)
(d) \( \dfrac{ehB}{16\pi m} \)
Show Answer & Explanation

Answer: (b) \( \dfrac{ehB}{8\pi m} \)

Explanation:
1. \( \tau=\mu B\sin\theta \) with \( \theta=30^\circ \).
2. \( \tau=\dfrac{eh}{4\pi m}B\cdot\dfrac12=\dfrac{ehB}{8\pi m} \).

Teacher's Note:
1. Express the quantity through the magnetic moment \( m=IA \) and the angular momentum \( L \), or use \( \vec\tau=\vec m\times\vec B \).
2. Cross-check by dimensions or a special case.

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