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Axis of a solid cylinder of infinite length and radius R lies along y – axis it carries a uniformly distributed current ‘I’ along + y direction. Magnetic field at a point $(\frac{R}{2}, y, \frac{R}{2})$ is

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a

$\frac{μ_{0} I}{4 πR} (\hat{i} - \hat{k})$

b

$\frac{μ_{0} I}{2 πR} (\hat{j} - \hat{k})$

c

$\frac{μ_{0} I}{4 πR} \hat{j}$

d

$\frac{μ_{0} I}{4 πR} (\hat{i} + \hat{k})$

answer is A.

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Detailed Solution

Magnitude of field at Point $P (\frac{R}{2}, y, \frac{R}{2})$ is $B = \frac{μ_{0} j r}{2} = \frac{μ_{0} I}{2 {πR}^{2}} \times \frac{R}{\sqrt{2}} = \frac{μ_{0} I}{2 \sqrt{2} πR}$
Unit vector in the direction of field, $\hat{B} = \frac{\hat{i} - \hat{k}}{\sqrt{2}}$

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