Question

Magnetic fields at two points on the axis of a circular coil at a distance of 0.05 m and 0.2 m from the centre are in the ratio 8 : 1. The radius of coil is _____________

• A. 0.1 m
• B. 0.15 m
• C. 0.2 m
• D. 1.0 m

Hint:

The magnetic field is maximum at the center ($x=0$) and decreases as you move away along the axis. The $3/2$ power in the denominator is key to these ratio problems.

Answer 1

The Correct Option is A

To determine the radius of the coil, we must understand the formula for the magnetic field along the axis of a circular coil. The magnetic field B at a distance x from the center of a coil of radius R carrying current I is given by:

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

where \mu_0 is the permeability of free space.

We have two distances, x_1 = 0.05 \, \text{m} and x_2 = 0.2 \, \text{m}. The magnetic fields at these points are in the ratio 8:1. Let B_1 and B_2 be the magnetic fields at these distances respectively.

The ratio of the magnetic fields is given by:

\frac{B_1}{B_2} = \frac{(R^2 + x_2^2)^{3/2}}{(R^2 + x_1^2)^{3/2}}

According to the question:

\frac{B_1}{B_2} = 8

Thus,

8 = \frac{(R^2 + (0.2)^2)^{3/2}}{(R^2 + (0.05)^2)^{3/2}}

This implies:

(R^2 + (0.2)^2)^{3/2} = 8 \times (R^2 + (0.05)^2)^{3/2}

To simplify, taking cube roots on both sides:

R^2 + (0.2)^2 = 8^{2/3} \times (R^2 + (0.05)^2)

Calculating 8^{2/3} \approx 4, we get:

R^2 + 0.04 = 4 \times (R^2 + 0.0025)

Expanding and simplifying further:

R^2 + 0.04 = 4R^2 + 0.01

Rearranging the terms:

3R^2 = 0.03

Solving for R^2:

R^2 = \frac{0.03}{3} = 0.01

Therefore, the radius R is:

R = \sqrt{0.01} = 0.1 \, \text{m}

Thus, the radius of the coil is 0.1 m.