Question

# The intensity of magnetic induction at the centre of a current-carrying circular coil is $$B_{1}$$ and at a point on its axis at a distance equal to its radius from the centre is $$B_{2}$$, then $$\dfrac{B_{1}}{B_{2}}$$ is

A
22
B
122
C
12
D
2

Solution

## The correct option is A $$2\sqrt{2}$$$$B_1=\dfrac {\mu_0 i}{2a}$$$$B_2=\dfrac {\mu_0\cdot i\cdot a^2}{2(a^2+a^2)^{3/2}}$$$$=\dfrac {\mu_0 ia^2}{2\times 2\sqrt 2\times a^3}$$$$=\dfrac {\mu_0 i}{4\sqrt 2\times a}$$$$\dfrac {B_1}{B_2} = 2\sqrt 2$$Physics

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