Question

A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction $\frac{1}{\pi}\big(\frac{Wb}{m^2}\big)$ in such a way that its axis makes an angle of 60$^{\circ}$ with $\overrightarrow{B}$ . The magnetic flux linked with the disc is

• A. 0.08 Wb
• B. 0.01 Wb
• C. 0.02 Wb
• D. 0.06 Wb

Answer 1

The Correct Option is C

To solve the problem of finding the magnetic flux linked with a circular disc in a uniform magnetic field, we will use the formula for magnetic flux, \(\Phi\), which is given by:

\[\Phi = B \cdot A \cdot \cos \theta\]

Where:

• \(B\) is the magnetic field induction.
• \(A\) is the area of the disc.
• \(\theta\) is the angle between the magnetic field and the normal to the surface of the disc.

Given:

• Radius of the disc, \(r = 0.2 \, \text{m}\).
• Magnetic field induction, \(B = \frac{1}{\pi} \, \text{Wb/m}^2\).
• Angle, \(\theta = 60^\circ\).

Firstly, let's find the area \(A\) of the circular disc:

\[A = \pi r^2 = \pi (0.2)^2 = \pi \times 0.04 = 0.04\pi \, \text{m}^2\]

Now, using the formula for magnetic flux:

\[\Phi = B \cdot A \cdot \cos \theta\]

Substitute the known values:

\[\Phi = \left(\frac{1}{\pi}\right) \cdot (0.04\pi) \cdot \cos 60^\circ\]

Keeling in mind that \(\cos 60^\circ = \frac{1}{2}\), proceed with calculation:

\[\Phi = \left(\frac{1}{\pi}\right) \cdot (0.04\pi) \cdot \frac{1}{2}\]

\[\Phi = 0.04 \cdot \frac{1}{2} = 0.02 \, \text{Wb}\]

The magnetic flux linked with the disc is 0.02 Wb, which matches the correct answer choice.