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A circular coil of, radius R, carries a current I. The magnetic field at the centre of the coil is B . Then B equals.
where. c is the speed of light in vacuum.

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$\frac{μ_{0} I}{2 πR}$

$\frac{I}{2 C^{2} ε_{0} R}$

$\frac{μ_{0} I}{2 R}$

$\frac{{Ic}^{2}}{2 ε_{0} R}$

answer is A, C.

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Detailed Solution

Magnetic field at the centre of circular coil is given by

$B = \frac{μ_{0} I}{2 R}$

Since we know that $c^{2} = \frac{1}{μ_{0} ε_{0}}$

$\Rightarrow μ_{0} = \frac{1}{c^{2} ε_{0}} \Rightarrow B = \frac{1}{2 c^{2} ε_{0} R}$

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