Question

The intensity of the magnetic field \( B \) due to a current-carrying circular coil of radius 12 cm at its center is \( 0.5 \times 10^{-4} \, \text{Tesla} \) perpendicular to the plane of the coil upward. Calculate the magnitude and direction of current flowing in the coil. 
 

Hint:

The magnetic field at the center of a current-carrying circular coil is given by \( B = \frac{\mu_0 I}{2R} \), where \( \mu_0 \) is the permeability, \( I \) is the current, and \( R \) is the radius of the coil.

Answer 1

The magnetic field at the center of a circular loop carrying current \( I \) is given by: \[ B = \frac{\mu_0 I}{2R}, \] where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) is the permeability of free space, \( I \) is the current, and \( R \) is the radius of the coil. Rearranging for \( I \): \[ I = \frac{2 B R}{\mu_0}. \] Substituting the given values (\( B = 0.5 \times 10^{-4} \, \text{T} \), \( R = 12 \, \text{cm} = 0.12 \, \text{m} \)): \[ I = \frac{2 \times 0.5 \times 10^{-4} \times 0.12}{4\pi \times 10^{-7}} = 3.0 \, \text{A}. \] The direction of the current follows the right-hand rule: when the fingers of the right hand curl in the direction of the current in the coil, the thumb points in the direction of the magnetic field at the center, which is upward.