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The magnetic field at a distance x on the axis of a circular coil of radius R is 18th of that of the centre. The value of x is
A) R3
B)2R3
C)R3
D)R2

Answer
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Hint: In the above question the magnetic field along the axis of the coil is at its centre is given to us. The magnetic field along the axis is governed by a particular relation. Hence comparing the magnetic field at the centre of the coil and at distance x such that the magnetic field is 18th of that of the centre, will enable us to determine the required value of x.

Formula used:
B=μiR22(x2+R2)3/2

Complete step-by-step answer:
Let us say we have a circular coil of radius R carrying current ‘i’ as shown in the figure below.
seo images


The magnetic field at a distance x from the centre of the coil is given by,
B=μiR22(x2+R2)3/2 where μ is the permeability of free space.
At x = 0, i.e. the centre of the coil the magnetic field is equal to,
B=μiR22(x2+R2)3/2B=μiR22((0)2+R2)3/2B=μi2R.....(1)
Similarly the magnetic field at a distance x from the centre is equal to,
B1=μiR22(x2+R2)3/2.....(2)
In the question it is given that the magnetic field at a distance x on the axis of a circular coil of radius R is 18th of that of the centre. Therefore taking the ratio of equation 1 and 2 we get,
B1B=μiR22(x2+R2)3/2μi2RB1B=R3(x2+R2)3/2B1=18BB18B1=R3(x2+R2)3/28R3=(x2+R2)3/22R=(x2+R2)1/24R2=x2+R2x2=3R2x=3R

So, the correct answer is “Option C”.

Note: It is to be noted that we have taken the cube root on both the sides of the equation so that the power 3 on both the sides becomes 1. Further squaring on both the sides will help to get rid of the square root in the subsequent step. In the above question we have assumed the coil to be of single loop as the value of magnetic field in case of comparison does not matter.