Question

The number of turns of two circular coils A and B are 300 and 200 respectively. The magnetic moments of the two coils A and B are in the ratio 1:2. If the two coils carry equal currents, then the ratio of radii of coils A and B is

• A. 2:\(\sqrt{3}\)
• B. 2:3
• C. 1:2
• D. 1:\(\sqrt{3}\)

Hint:

When dealing with ratios, set up the formula for the quantity in question and then write the ratio of the two cases. Cancel out any quantities that are constant or given to be equal. This simplifies the problem to an algebraic equation for the unknown ratio.

Answer 1

The Correct Option is D

The magnetic moment (\(M\)) of a circular coil is given by the formula \( M = nIA \), where n is the number of turns, I is the current, and A is the area of the coil.
The area of a circular coil with radius r is \( A = \pi r^2 \).
So, the magnetic moment is \( M = nI(\pi r^2) \).
Let's write down the information given for coil A and coil B.
For coil A: \( n_A = 300 \), \( M_A \), \( I_A \), \( r_A \).
For coil B: \( n_B = 200 \), \( M_B \), \( I_B \), \( r_B \).
We are given that the currents are equal: \( I_A = I_B = I \).
We are given the ratio of their magnetic moments: \( \frac{M_A}{M_B} = \frac{1}{2} \).
Now, let's write the ratio of the magnetic moments using the formula.
\( \frac{M_A}{M_B} = \frac{n_A I_A (\pi r_A^2)}{n_B I_B (\pi r_B^2)} \).
Since \(I_A = I_B\) and \(\pi\) is a constant, they cancel out.
\( \frac{M_A}{M_B} = \frac{n_A r_A^2}{n_B r_B^2} \).
We are given this ratio is 1/2. Substitute the known values.
\( \frac{1}{2} = \frac{300 \cdot r_A^2}{200 \cdot r_B^2} = \frac{3}{2} \left( \frac{r_A}{r_B} \right)^2 \).
Now, we need to solve for the ratio of the radii, \( \frac{r_A}{r_B} \).
\( \left( \frac{r_A}{r_B} \right)^2 = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \).
Taking the square root of both sides:
\( \frac{r_A}{r_B} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \).
The ratio of the radii of coils A and B is \( 1:\sqrt{3} \).