Question

Let \( B_1 \) be the magnitude of magnetic field at the center of a circular coil of radius \( R \) carrying current \( I \). Let \( B_2 \) be the magnitude of magnetic field at an axial distance \( x \) from the center. For \( x : R = 3 : 4 \), \( \frac{B_2}{B_1} \) is:

• A. 4 : 5
• B. 16 : 25
• C. 64 : 125
• D. 25 : 16

Hint:

When dealing with the magnetic field produced by a circular coil, remember that the magnetic field is strongest at the center and weakens with distance along the axis.

Answer 1

The Correct Option is C

The magnetic field at the center of a circular coil is calculated as: \[ B_1 = \frac{\mu_0 I}{2R} \] The magnetic field at an axial distance \( x \) from the center is determined by: \[ B_2 = B_1 \sin \theta = \frac{B_1 \cdot \left( \frac{R}{x} \right)^2}{5} \] Given the ratio \( x : R = 3 : 4 \), the resultant magnetic field ratio is: \[ \frac{B_2}{B_1} = \frac{64}{125} \]