Question

When a certain length of wire is turned into one circular loop, the magnetic induction at the centre of the coil due to current \(I\) is \(B_1\). If the same wire is turned into four loops to make a circular coil, the magnetic induction at the centre of this coil is \(B_2\) for the same current. Then the relation between \(B_2\) and \(B_1\) is

• A. \( B_2 = 8B_1 \)
• B. \( B_2 = 64B_1 \)
• C. \( B_2 = 4B_1 \)
• D. \( B_2 = 16B_1 \)

Hint:

Magnetic field of a circular coil depends on both number of turns and radius.

Answer 1

The Correct Option is D

Step 1: Magnetic field at the centre of a circular coil.
For a coil of radius \(R\) with \(n\) turns carrying current \(I\):
\[ B = \frac{\mu_0 n I}{2R} \]
Step 2: Case of one loop.
Let the total length of wire be \(L\). Then,
\[ 2\pi R_1 = L \Rightarrow R_1 = \frac{L}{2\pi} \] \[ B_1 = \frac{\mu_0 I}{2R_1} \]
Step 3: Case of four loops.
Each loop has length \(L/4\):
\[ 2\pi R_2 = \frac{L}{4} \Rightarrow R_2 = \frac{L}{8\pi} \] Number of turns \(n = 4\).
Step 4: Magnetic field for four loops.
\[ B_2 = \frac{\mu_0 \times 4I}{2R_2} = \frac{2\mu_0 I}{R_2} \]
Step 5: Taking ratio of fields.
\[ \frac{B_2}{B_1} = \frac{\frac{2\mu_0 I}{R_2}}{\frac{\mu_0 I}{2R_1}} = 4 \times \frac{R_1}{R_2} \]
Step 6: Substituting radii.
\[ \frac{R_1}{R_2} = \frac{L/(2\pi)}{L/(8\pi)} = 4 \]
Step 7: Final result.
\[ \frac{B_2}{B_1} = 4 \times 4 = 16 \Rightarrow B_2 = 16B_1 \]