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A circular coil of radius $R$ carries current $I$. The magnetic field at the centre of the coil is:
A$B = \dfrac{\mu_0 I}{2\pi R}$
B$B = \mu_0 I R$
C$B = \dfrac{\mu_0 I}{2 R}$
D$B = \dfrac{\mu_0 I}{4\pi R}$
Answer & Solution
Correct answer: C. $B = \dfrac{\mu_0 I}{2 R}$
From Biot-Savart's law applied to a circular loop of radius $R$ carrying current $I$, the field at the loop's centre is
$B = \dfrac{\mu_0 I}{2 R}$.
Direction along the axis of the loop. Right-hand rule again — curl fingers along current, thumb gives field direction.
The field along the axis at distance $x$ from the centre is $B(x) = \dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$, which collapses to the centre formula when $x = 0$.
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