Paragraph: A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity ω . This can be considered as equivalent to a loop carrying a steady current (Q ω)/(2 π) . A uniform

Correct answer: B. - γ BQR ^ 2 2 Magnetic dipole moment M = γ J Δ M = γ Δ J ...(1) Δ J Δ t = - Q dB dt · R 2 R Δ J = - QB 2 R ^ 2 So Δ M = - γ QBR ^ 2 2 Alternate M L = Q 2m M = Qω 2π π R ^ 2 = Qω R ^ 2 2 induced electric field is opposite to the ω so the charge is retarded. ω ^ ′ = ω - α t ω ^ ′ = ω - QB 2 1 ( a _ t = QE/m) , (α = QE mR = Q R × BR 2 m = QB 2 m ) M _ f = Q ω ^ ′ R ^ 2 2 = Q (ω - QB 2 m ) R ^ 2 2 ∆ M = M _ f - M _ i = Qω R ^ 2 2 - Q ^ 2 BR ^ 2 4m - Qω R ^ 2 2 = - γ BQR ^ 2 2

JEE Advanced 2013Paper 2

Exam: JEE Advanced
Year: 2013
Paper: Paper 2
Subject: Physics
Chapter: Electromagnetic Induction

JEE Advanced 2013 · Paper 2 · Physics · Electromagnetic Induction

Paragraph: A point charge Q is moving in a circular orbit of radius...

Paragraph: A point charge

Q

is moving in a circular orbit of radius

R

in the

x-y

plane with an angular velocity

\omega .

This can be considered as equivalent to a loop carrying a steady current

\frac{Q \omega}{2 \pi}

. 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

\gamma

.
Question: The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is Paragraph: A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity ω .

Paragraph: A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity ω .

Options

A $- \gamma {\text{BQR}}^{2}$
B $- \gamma \dfrac{{\text{BQR}}^{2}}{2}$
C $\gamma \dfrac{{\text{BQR}}^{2}}{2}$
D $\gamma {\text{BQR}}^{2}$

Correct answer

B. $- \gamma \dfrac{{\text{BQR}}^{2}}{2}$

Step-by-step solution

Magnetic dipole moment

\text{M} = \gamma \text{J}

\Delta \text{M} = \gamma \Delta \text{J}

...(1)

\dfrac{\Delta \text{J}}{\Delta \text{t}} = - \text{Q} \dfrac{\text{dB}}{\text{dt}} \cdot \dfrac{\text{R}}{2} \text{R}

\Delta \text{J} = - \dfrac{\text{QB}}{2} {\text{R}}^{2}

\Delta \text{M} = - \dfrac{\gamma {\text{QBR}}^{2}}{2}

Alternate

\dfrac{\text{M}}{\text{L}} = \dfrac{\text{Q}}{2\text{m}}

\text{M} = \dfrac{\text{Q}\omega}{2\pi}\pi{\text{R}}^{2} = \dfrac{\text{Q}\omega{\text{R}}^{2}}{2}

induced electric field is opposite to the

\omega

so the charge is retarded.

{\omega}^{′} = \omega - \alpha \text{t}

{\omega}^{′} = \omega - \dfrac{\text{QB}}{2} 1 \left({\text{a}}_{\text{t}} = \text{QE/m}\right) \text{,} \left(\alpha = \dfrac{\text{QE}}{\text{mR}} = \dfrac{\text{Q}}{\text{R}} \times \dfrac{\text{BR}}{2 \text{m}} = \dfrac{\text{QB}}{2 \text{m}}\right)

{\text{M}}_{\text{f}} = \dfrac{\text{Q} {\omega}^{′} {\text{R}}^{2}}{2} = \text{Q} \left(\omega - \dfrac{\text{QB}}{2 \text{m}}\right) \dfrac{{\text{R}}^{2}}{2}

∆ \text{M} = {\text{M}}_{\text{f}} - {\text{M}}_{\text{i}} = \dfrac{\text{Q}\omega{\text{R}}^{2}}{2} - \dfrac{{\text{Q}}^{2}{\text{BR}}^{2}}{4\text{m}} - \dfrac{\text{Q}\omega{\text{R}}^{2}}{2} = - \gamma\dfrac{{\text{BQR}}^{2}}{2}

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