JEE Main · 2004 · 24 Apr 2004 · Offline · Magnetic Effects of Current

The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is

54 μ T

. What will be its value at the centre of the loop?

Answer: (A)

Solution

On the axis of a circular loop the field is

B = \frac{μ _{0} i R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}}

. With

R = 3

cm and

x = 4

cm,

(R^{2} + x^{2})^{3/2} = 2 5^{3/2} = 125

, so the axial reading of

54 μ

T fixes

μ_{0} i

. At the centre, where

x = 0

, the field is

\frac{μ _{0} i}{2 R}

, so the ratio is

\frac{B _{ce n t r e}}{B _{a x i s}} = \frac{( R ^{2} + x ^{2} ) ^{3/2}}{R ^{3}} = \frac{125}{27}

. Hence

B_{ce n t r e} = 54 \times \frac{125}{27} = 250 μ

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option