Search any question & find its solution
Question:
Answered & Verified by Expert
A circular disc of radius $0.2 \mathrm{~m}$ is placed in a uniform magnetic field of induction $\frac{1}{\pi}\left(\frac{W b}{m^2}\right)$ in such a way that its axis makes an angle of $60^{\circ}$ with $\overrightarrow{\mathbf{B}}$. The magnetic flux linked with the disc is
Options:
Solution:
2336 Upvotes
Verified Answer
The correct answer is:
$0.02 \mathrm{~Wb}$
The magnetic flux $\phi$ passing surface of area $A$ placed in a u field $B$ is given by
$\phi=B A \cos \theta$
where $\theta$ is the angle between
$\begin{aligned}
& \text {Here, } \theta=60^{\circ}, B=\frac{1}{\pi} \mathrm{Wb} / \mathrm{m}^2, A=\pi(0.2)^2 \\
& \text {Therefore, } \phi=\frac{1}{\pi} \times \pi(0.2)^2 \times \cos 60^{\circ} \\
& =(0.2)^2 \times \frac{1}{2} \\
& =0.02 \mathrm{~Wb} \\
&
\end{aligned}$
$\phi=B A \cos \theta$
where $\theta$ is the angle between
$\begin{aligned}
& \text {Here, } \theta=60^{\circ}, B=\frac{1}{\pi} \mathrm{Wb} / \mathrm{m}^2, A=\pi(0.2)^2 \\
& \text {Therefore, } \phi=\frac{1}{\pi} \times \pi(0.2)^2 \times \cos 60^{\circ} \\
& =(0.2)^2 \times \frac{1}{2} \\
& =0.02 \mathrm{~Wb} \\
&
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.