From a cylinder of radius R, a cylinder of radius R/2 is removed, as shown in Fig. Current flowing in the remaining cylinder is I. Then, magnetic field strength is

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$\frac{μ_{0} I}{2 π R}$ at point A

$\frac{μ_{0} I}{3 π R}$ at point B

zero at point A

zero at point B

answer is D.

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Detailed Solution

Current density $j = \frac{I}{π R^{2} - π {(R / 2)}^{2}} = \frac{4 I}{3 π R^{2}}$

Current in the smaller cylinder is $I_{s m a l l} = j . π {(R / 2)}^{2} = \frac{I}{3}$

For A:- $B_{A} = 0 - \frac{μ_{0} (I / 3)}{2 π (R / 2)}$ , For B:- $B_{B} = \frac{μ_{0} I}{3 π R}$

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