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Electricity and Magnetism Level 4

The magnetic induction due to circular current carrying conductor of radius $a$ , at the center is $B_c$ . The magnetic induction on its axis at a distance $a$ from its centre is $B_a$ . Find $\dfrac{B_c} {B_a}$ .

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Vishwak Srinivasan
Jun 22, 2015

Magnetic field induction due to a circular current carrying coil is at a distance $x$ from the centre along the axis is:

$B_x = \dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{\frac{3}{2}}}$

At centre:

$B_c = B_0 = \dfrac{\mu_0 I}{2 a}$

At a distance $a$ :

$B_a = \dfrac{\mu_0 I}{4\sqrt{2} a}$

$\dfrac{B_c}{B_a} = \boxed{2\sqrt{2}}$

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The answer is 2.82.

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