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# For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given

16.(a) For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,

$B = \frac{\mu _0 IR^2N}{2 ( x^2 + R^2 )^{3/2}}$

Show that this reduces to the familiar result for the field at the centre of the coil.

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For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,

$B = \frac{\mu _0 IR^2N}{2 ( x^2 + R^2 )^{3/2}}$

For finding the field at the centre of coil we put x=0 and get the familiar result

$B = \frac{\mu _0 IN}{2R}$

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