GATE 2020 PH – Question 43
A uniform field $\mathbf B=B_0\hat{\mathbf y}$ exists in frame K. A perfectly conducting sphere moves at $\mathbf v=v_0\hat{\mathbf x}$; its rest frame is K\prime. The field transformations are $E_\parallel\prime=E_\parallel$, $\mathbf E_\perp\prime=\gamma(\mathbf E_\perp+\mathbf v\times\mathbf B)$, $B_\parallel\prime=B_\parallel$, $\mathbf B_\perp\prime=\gamma(\mathbf B_\perp-\mathbf v\times\mathbf E/c^2)$, with $\gamma=(1-v^2/c^2)^{-1/2}$. To lowest order in v/c, where is the induced surface charge density greatest in magnitude in K\prime?
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Correct answer: (A) Along z′.
Explanation
In K the imposed electric field is zero. To first order, the sphere’s rest frame sees $\mathbf E\prime\simeq\mathbf v\times\mathbf B=v_0B_0\hat{\mathbf z}\prime$. Charges redistribute to cancel the field inside the conductor. For a sphere in a uniform field, the induced surface density varies as the cosine of angle to that field, with extrema at its poles. Therefore the largest magnitude is along the z′ axis, A.