Question

The magnetic field due to a current carrying circular loop of radius 4 cm at a point on the axis at a distance of 7 cm from the center is 48 μT. What will be the value at the center of the loop?

(a) 390 μT

(b) 393 μT

(c) 395 μT

(d) 397 μT

This question was addressed to me during an online interview.

The question is from Magnetic Field on the Axis of a Circular Current Loop topic in section Moving Charges and Magnetism of Physics – Class 12

Answer 1

Correct answer is (b) 393 μT

For explanation: Field along axis of coil → B =\( \frac {\mu_o IR^2}{2 \big ( (R^2 \, + \, x^2)^{\frac {3}{2}} \big ] } \)

At the coil of the coil → B1 = \( \frac {\mu_o I}{2R}\)

\(\frac {B_1}{B} = \frac {\mu_o I}{2R} \, \times \, \frac {2 \big ( (R^2 \, + \, x^2)^{\frac {3}{2}} \big ] }{\mu_o IR^2} = \frac {2 \big ( (R^2 \, + \, x^2)^{\frac {3}{2}} \big ] }{R^3}\)

B1 = \(\frac {48 [4^2 + 7^2]^{\frac {3}{2}}}{4^3}\)

B1 = \(\frac {48 \times (65)^{\frac {3}{2}}}{4^3}\)

B1 = 131.01 × 3

B1 = 393.03 = 393 μT