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Question

The magnetic induction at a point at a large distance d on the axial line of circular coil of small radius carrying current is 120 $$\mu T$$. At a distance 2d the magnetic induction would be:


A
60μT
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B
30μT
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C
15μT
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D
240μT
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Solution

The correct option is C 15$$\mu T$$
The magnetic field at a point at a distance d on the axial line of circular coil of radius r is given by , $$B=\dfrac{\mu_0Ir^2}{2(r^2+d^2)^{3/2}}$$ 
So, $$B=\dfrac{\mu_0Ir^2}{2[d^2(1+r^2/d^2)]^{3/2}}=\dfrac{\mu_0Ir^2}{2d^3}(1+r^2/d^2)^{-3/2}=\dfrac{\mu_0Ir^2}{2d^3}(1-3a^2/2d^2+ ...)$$
We get   $$B=\dfrac{\mu_0Ir^2}{2d^3}$$ 
(As $$d>>r$$ so higher order term can be negligible)
Thus, $$\dfrac{B_{2d}}{B_d}=\dfrac{d^3}{(2d)^3}$$
or $$B_{2d}=B_d/8=120/8=15\mu T$$

Physics

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