Question

A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction $\dfrac{1}{\pi }\left( {{\text{Wb}}{{\text{m}}^{ - 2}}} \right)$ in such a way that its axis makes an angle of $60^\circ$ with $\vec B$. The magnetic flux linked with the disc is:
A. $0.08\,{\text{Wb}}$
B. $0.01\,{\text{Wb}}$
C. $0.02\,{\text{Wb}}$
D. $0.06\,{\text{Wb}}$

Answer 1

First of all, we will find the area of the disc and the component of the magnetic field required to find the magnetic flux. We will substitute the values and manipulate accordingly.

Complete step by step solution:
In the given problem,
The radius of the circular disc is $0.2$ meter.
The uniform magnetic field of induction is $\dfrac{1}{\pi }\left( {{\text{Wb}}{{\text{m}}^{ - 2}}} \right)$ .
The axis of the disc is inclined to the magnetic field with an angle of $60^\circ$ .
We need to find the magnetic flux linked with the disc.
For this, we have a formula, which gives magnetic flux linked with it:
$\phi = {\rm B}{\rm A}\cos \theta$ …… (1)
Where,
$\phi$ indicates magnetic flux linked with the disc.
${\rm B}$ indicates magnetic field.
${\rm A}$ indicates the area of the disc.
$\theta$ indicates the inclination of the axis of the disc to the magnetic field.
In the formula, we use the cosine component of the magnetic field, because we need to find the magnetic flux, i.e. it means we need to find the total number of magnetic field lines which passes through the given area. The component which lies along the axis of the disc is always the cosine component.
The radius of the circular disc is $0.2$ meter.
We can calculate its area, given by the formula:

$$A = \pi {r^2} \\\ \Rightarrow A = \pi \times {\left( {0.2} \right)^2}\,{{\text{m}}^2}$$

Now, substituting the required values in the equation (1), we get:

$$\phi = {\rm B}{\rm A}\cos \theta \\\ \Rightarrow \phi = \dfrac{1} {\pi } \times \pi \times {\left( {0.2} \right)^2} \times \cos 60^\circ \\\ \Rightarrow \phi = 0.04 \times \dfrac{1} {2} \\\ \Rightarrow \phi = 0.02\,{\text{Wb}}\,$$

Hence, the magnetic flux linked to with the disc is $0.02\,{\text{Wb}}$.

**The correct option is C. $0.02\,{\text{Wb}}$

Note: **
This problem can be solved if we have some knowledge on magnetic flux and magnetic fields. While solving the problem, don’t use the sine component of the magnetic field, as this component indicates the magnetic field, perpendicular to the axis of disc. Since, we are asked to find the magnetic flux (number of field lines passing perpendicular to the plane of the disc), we use the cosine component.