JEE Challenger
More from Electromagnetic Induction
PhyHardAdv 2013 • Paper 2

Change in Magnetic Dipole Moment of Orbiting Charge

Comprehension Passage

Consider a point charge QQ moving along a circular orbit with radius RR within the x−yx-y plane at an angular velocity ω\omega. This setup may be treated as equivalent to a current loop carrying a steady current Qω2π\frac{Q\omega}{2\pi}. A uniform magnetic field parallel to the positive zz-axis is then turned on, increasing at a uniform rate from 00 to BB over a duration of one second. Assume the orbit's radius stays constant. Applying this magnetic field induces an electromotive force (emf) along the orbit, defined as the work performed by the induced electric field to move a unit positive charge along the loop. Furthermore, for an orbiting charge, its magnetic dipole moment is directly proportional to its angular momentum via a proportionality constant γ\gamma.

Determine the change in the magnetic dipole moment associated with the orbit at the conclusion of the time interval during which the magnetic field changes.

Options

A

−γBQR2-\gamma BQR^2

B

−γBQR22-\gamma \frac{BQR^2}{2}

Correct
C

γBQR22\gamma \frac{BQR^2}{2}

D

γBQR2\gamma BQR^2

Topics & Concepts

Electromagnetic InductionFaraday's LawAngular Momentum

Step-by-Step Solution

To find the change in the magnetic dipole moment of the orbiting charge, we analyze the effect of the induced electric field generated by the changing magnetic field.

1. Induced Electric Field: According to Faraday's law of electromagnetic induction, the induced electromotive force (emf) around the circular orbit of radius RR is given by: E=−dΦBdt=−ddt(B⋅πR2)=−πR2dBdt\mathcal{E} = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}\left(B \cdot \pi R^2\right) = -\pi R^2 \frac{dB}{dt}

Since the orbit is symmetric, the induced electric field EE is uniform along the circular path: ∮E⃗⋅dl⃗=E(2πR)=−πR2dBdt\oint \vec{E} \cdot d\vec{l} = E(2\pi R) = -\pi R^2 \frac{dB}{dt} E=−R2dBdtE = -\frac{R}{2}\frac{dB}{dt}

2. Torque and Angular Momentum: The tangential force exerted on the charge QQ by this induced electric field is: F=QE=−QR2dBdtF = Q E = -\frac{Q R}{2}\frac{dB}{dt}

The torque τ\tau acting on the charge about the center of the orbit is: τ=RF=−QR22dBdt\tau = R F = -\frac{Q R^2}{2}\frac{dB}{dt}

Using the relation between torque and the rate of change of angular momentum (τ=dLdt\tau = \frac{dL}{dt}): dLdt=−QR22dBdt\frac{dL}{dt} = -\frac{Q R^2}{2}\frac{dB}{dt}

Integrating both sides over the duration during which the magnetic field changes from 00 to BB: ΔL=∫dL=−QR22∫0BdB=−BQR22\Delta L = \int dL = -\frac{Q R^2}{2} \int_0^B dB = -\frac{B Q R^2}{2}

3. Change in Magnetic Dipole Moment: It is given that the magnetic dipole moment μ\mu is directly proportional to the angular momentum LL with proportionality constant γ\gamma: μ=γL\mu = \gamma L

Therefore, the change in the magnetic dipole moment Δμ\Delta \mu is: Δμ=γΔL=−γBQR22\Delta \mu = \gamma \Delta L = -\gamma \frac{B Q R^2}{2}

Thus, the correct choice is B.

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