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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Advanced 2025 · Paper 2 · Physics · Magnetic Effects of Current

A conducting solid sphere of radius R and mass M carries a charge Q...

A conducting solid sphere of radius RR and mass MM carries a charge QQ. The sphere is rotating about an axis passing through its center with a uniform angular speed ω\omega. The ratio of the magnitudes of the magnetic dipole moment to the angular momentum about the same axis is given as αQ2M\alpha \frac{Q}{2 M}. The value of α\alpha is _______ .A conducting solid sphere of radius R and mass M carries a charge Q . The sphere is rotating about an axis passing

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A conducting solid sphere of radius R and mass M carries a charge Q . The sphere is rotating about an axis passing
dM=dIAA=πr2=π(Rsin⁡θ)2dI=daT=σ(2πr)(Rdθ)ω2πdI=σ2πR2ωsin⁡θdθ2π\begin{aligned} & d M=d I A \\ & A=\pi r^2=\pi(R \sin \theta)^2 \\ & d I=\frac{d a}{T}=\frac{\sigma(2 \pi r)(R d \theta) \omega}{2 \pi} \\ & d I=\frac{\sigma 2 \pi R^2 \omega \sin \theta d \theta}{2 \pi}\end{aligned} dI=σR2ωsin⁡θ dθ\mathrm{dI}=\sigma \mathrm{R}^2 \omega \sin \theta \mathrm{~d} \theta Magnetic dipole moment : M=∫dM=∫0πσR2ωπR2sin⁡3θdθM=\int d M=\int_0^\pi \sigma R^2 \omega \pi R^2 \sin ^3 \theta d \theta M=σR4ωπ∫0πsin⁡3θdθM=\sigma R^4 \omega \pi \int_0^\pi \sin ^3 \theta d \theta \quad\left(\because \int_0^\pi \sin ^3 \theta d \theta=\frac{4}{3}\right)M=(Q4πR2)R4ωπ(43)\mathrm{M}=\left(\frac{\mathrm{Q}}{4 \pi \mathrm{R}^2}\right) \mathrm{R}^4 \omega \pi\left(\frac{4}{3}\right) Magnetic dipole moment M=QR2ω3\mathrm{M}=\frac{\mathrm{QR}^2 \omega}{3} Angular momentum L=(25MR2)ω\mathrm{L}=\left(\frac{2}{5} \mathrm{MR}^2\right) \omega M L=QR2ω3×25MR2ω=Q2M(53)\frac{\mathrm{M}}{\mathrm{~L}}=\frac{\mathrm{QR}^2 \omega}{3 \times \frac{2}{5} \mathrm{MR}^2 \omega}=\frac{\mathrm{Q}}{2 \mathrm{M}}\left(\frac{5}{3}\right)\alpha=\frac{5}{3}=1.67
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