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 defined as \( \Phi_B = B A \cos \theta \). Here, \( B \) represents the magnetic field strength, \( A \) is the coil's area, and \( \theta \) is the angle between the magnetic field and the normal to the coil's surface. The coil's area is calculated as \( A = \pi r^2 \). Given a radius \( r = 0.1 \) m, the area is \( A = \pi (0.1)^2 = 0.0314 \, \text{m}^2 \). The magnetic field \( B \) is determined by the dot product of the magnetic field vector \( \vec{B} \) and the unit normal vector \( \hat{n} \) to the coil's surface: \( B = \vec{B} \cdot \hat{n} \). For \( \vec{B} = 1.0 \hat{i} + 0.5 \hat{j} \) and \( \hat{n} = 0.6 \hat{i} + 0.8 \hat{j} \), the magnetic field strength is \( B = (1.0 \hat{i} + 0.5 \hat{j}) \cdot (0.6 \hat{i} + 0.8 \hat{j}) = 1.0 \times 0.6 + 0.5 \times 0.8 = 0.6 + 0.4 = 1.0 \, \text{mT} \). Consequently, the magnetic flux is computed as \( \Phi_B = B A = 1.0 \times 0.0314 = 0.0314 \, \text{Wb} \), which is equivalent to \( 31.4 \, \mu \text{Wb} \).