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Magnetism question

2013 · Shift 2 · Q55
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Magnetism question

2013 · Shift 2 · Q55

JEE AdvancedPhysicsMagnetismMCQ+3 / −1
A point charge Q is moving in a circular orbit of radius R in the xy-plane with an angular velocity ω\omegaω. This can be considered as equivalent to a loop carrying a steady current Qω2π{{Q\omega } \over {2\pi }}2πQω​. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant γ\gammaγ.The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is
  1. A
    −γBQR2- \gamma BQ{R^2}−γBQR2
  2. B
    −γBQR22- \gamma {{BQ{R^2}} \over 2}−γ2BQR2​
  3. C
    γBQR22\gamma {{BQ{R^2}} \over 2}γ2BQR2​
  4. D
    γBQR2\gamma BQ{R^2}γBQR2
View written solutionFree

Correct answer: B

  1. Magnetic field changing with time induces an emf

A uniform magnetic field along +z+z+z increases from 000 to BBB in 111 s, so

rac{dB}{dt}=B \, .

For a circular orbit of radius RRR in the xyxyxy-plane, the magnetic flux is

Φ=BπR2.\Phi = B\pi R^2.Φ=BπR2.

Hence the induced emf is

E=−dΦdt=−πR2dBdt.\mathcal E = -\frac{d\Phi}{dt} = -\pi R^2\frac{dB}{dt}.E=−dtdΦ​=−πR2dtdB​.

So here,

E=−πR2B.\mathcal E = -\pi R^2 B.E=−πR2B.

  1. Relate emf to change in kinetic energy / angular speed

The induced emf is the work done per unit charge around the orbit. Therefore, for charge QQQ, the work done in 1 second is

W=QE=−QπR2B.W = Q\mathcal E = -Q\pi R^2 B.W=QE=−QπR2B.

This changes the kinetic energy of the orbiting charge.

Since radius remains constant,

K=12mv2=12m(ωR)2=12mR2ω2.K = \frac{1}{2}mv^2 = \frac{1}{2}m(\omega R)^2 = \frac{1}{2}mR^2\omega^2.K=21​mv2=21​m(ωR)2=21​mR2ω2.

Angular momentum is

L=mR2ω.L = mR^2\omega.L=mR2ω.

Thus,

K=L22mR2.K = \frac{L^2}{2mR^2}.K=2mR2L2​.

Differentiating,

dK=LmR2 dL=ω dL.dK = \frac{L}{mR^2}\, dL = \omega \, dL.dK=mR2L​dL=ωdL.

But a simpler route is through torque.

  1. Use induced electric field to find torque

Let the induced tangential electric field be EEE. Then

∮E⃗⋅dl⃗=E(2πR)=E.\oint \vec E\cdot d\vec l = E(2\pi R)=\mathcal E.∮E⋅dl=E(2πR)=E.

So,

Tangential force on the charge is

Ft=QE.F_t = QE.Ft​=QE.

Hence torque about the center is

τ=RFt=RQE=RQE2πR=QE2π.\tau = RF_t = RQE = RQ\frac{\mathcal E}{2\pi R} = \frac{Q\mathcal E}{2\pi}.τ=RFt​=RQE=RQ2πRE​=2πQE​.

Now,

τ=dLdt.\tau = \frac{dL}{dt}.τ=dtdL​.

Therefore,

= -\frac{Q R^2}{2}\frac{dB}{dt}.$$ Integrating from $B=0$ to $B=B$, $$\Delta L = -\frac{Q R^2}{2}\int_0^1 \frac{dB}{dt}\,dt = -\frac{Q R^2}{2}(B-0).$$ Thus, $$\Delta L = -\frac{BQ R^2}{2}.$$ 4. **Relate magnetic dipole moment to angular momentum** Given $$\mu = \gamma L.$$ Therefore, $$\Delta \mu = \gamma \, \Delta L = \gamma\left(-\frac{BQ R^2}{2}\right).$$ So, $$\boxed{\Delta \mu = -\gamma \frac{BQ R^2}{2}}.$$ 5. **Match with options** This corresponds to: $$\boxed{\text{Option B}}$$
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