The magnetic field due to a current carrying circular loop of radius 3cm at a point on the axis at a distance of 4cm from the centre is 54\,\mu T. What will be its value at the center of loop?

Magnetics PYQ — JEE Main 2004 · Shift 0 · Q143 with Solution

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Correct answer: C

For a loop of radius $R$ carrying current $I$ , the magnetic field at a point on its axis at distance $x$ from the center is

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

At the center of the loop $(x=0)$ ,

B_0 = \frac{\mu_0 I}{2R}

Given:

Now,

\frac{B_0}{B_x} = \frac{\mu_0 I/(2R)}{\mu_0 I R^2/[2(R^2+x^2)^{3/2}]}

Simplifying,

\frac{B_0}{B_x} = \frac{(R^2+x^2)^{3/2}}{R^3}

R^2+x^2 = 3^2+4^2 = 9+16=25

(R^2+x^2)^{3/2} = 25^{3/2} = (\sqrt{25})^3 = 5^3 = 125

R^3 = 3^3 = 27

So,

\frac{B_0}{B_x} = \frac{125}{27}

Hence,

B_0 = 54 \times \frac{125}{27}

B_0 = 2 \times 125 = 250\,\mu T

\boxed{250\,\mu T}

So the correct option is C.