Question

A coil of area 5 cm² having 20 turns is placed in a uniform magnetic field of \( 10^3 \, \text{gauss} \). The normal to the plane of coil makes an angle 30° with the magnetic field. The flux through the coil is

• A. \( 6.67 \times 10^{-4} \, \text{wb} \)
• B. \( 3.2 \times 10^{-5} \, \text{wb} \)
• C. \( 5.9 \times 10^{-4} \, \text{wb} \)
• D. \( 8.65 \times 10^{-4} \, \text{wb} \)

Hint:

The magnetic flux is given by the product of magnetic field strength, area, and the cosine of the angle between the normal to the coil and the magnetic field.

Answer 1

The Correct Option is A

Using the formula for magnetic flux \( \Phi = B A \cos \theta \), where \( B = 10^3 \, \text{gauss} = 10^{-1} \, \text{T} \), \( A = 5 \, \text{cm}^2 = 5 \times 10^{-4} \, \text{m}^2 \), and \( \theta = 30^\circ \), we calculate the flux as \( 6.67 \times 10^{-4} \, \text{wb} \).