Question

A circular coil of radius 10 cm is placed in a magnetic field \( \vec{B} = (1.0 \hat{i} + 0.5 \hat{j}) \, \text{mT} \) such that the outward unit vector normal to the surface of the coil is \( (0.6 \hat{i} + 0.8 \hat{j}) \). The magnetic flux linked with the coil is:

• A. \( 0.314 \, \mu \text{Wb} \)
• B. \( 3.14 \, \mu \text{Wb} \)
• C. \( 31.4 \, \mu \text{Wb} \)
• D. \( 1.256 \, \mu \text{Wb} \)

Hint:

For calculating magnetic flux, remember to use the dot product to find the effective magnetic field component normal to the coil's surface.

Answer 1

The Correct Option is C

The magnetic flux \( \Phi_B \) linked with the coil is given by: \[ \Phi_B = B A \cos \theta. \] Where:
\( B \) is the magnetic field,
\( A \) is the area of the coil,
\( \theta \) is the angle between the normal to the surface of the coil and the magnetic field.
The area of the coil is: \[ A = \pi r^2 = \pi (0.1)^2 = 0.0314 \, \text{m}^2. \] The magnetic field \( B \) is the dot product of the vector \( \vec{B} \) and the unit vector normal to the surface of the coil: \[ \vec{B} \cdot \hat{n} = (1.0 \hat{i} + 0.5 \hat{j}) \cdot (0.6 \hat{i} + 0.8 \hat{j}) = 1.0 \times 0.6 + 0.5 \times 0.8 = 0.6 + 0.4 = 1.0 \, \text{mT}. \] Thus, the magnetic flux is: \[ \Phi_B = 1.0 \times 0.0314 = 0.0314 \, \text{Wb} = 31.4 \, \mu \text{Wb}. \]