Q108 / 120 · VITEEE 2015 · Physics
A coil of area $5 \mathrm{~cm}^{2}$ having 20 turns is placed in a uniform magnetic field of $10^{3}$ gauss. The normal to the plane of coil makes an angle $30^{\circ}$ with the magnetic field. The flux through the coil is
- Correct answerA$6.67 \times 10^{-4} \mathrm{wb}$
- B$5.9 \times 10^{-4} \mathrm{wb}$
- C$3.2 \times 10^{-5} \mathrm{~Wb}$
- D$8.65 \times 10^{-4} \mathrm{wb}$
Solution
Given : $\mathrm{N}=20$
$\begin{array}{l}
\mathrm{B}=10^{3} \text { gauss } \\
=10^{3} \times 10^{-4} \mathrm{~T}=0.1 \mathrm{~T} \\
\mathrm{~A}=5 \mathrm{~cm}^{2}=5 \times 10^{-4} \mathrm{~m}^{2} \\
\theta=80^{\circ}
\end{array}$
$\therefore \quad$ Flux through the coil
$\begin{array}{l}
\begin{array}{l}
\phi=\text { NBA } \cos \theta \\
=20 \times 0.1 \times 5 \times 10^{-4} \times \cos 30^{\circ}
\end{array} \\
=10 \times 10^{-4} \times \frac{\sqrt{3}}{2}=5 \sqrt{3} \times 10^{-4}=865 \times 10^{-4} \mathrm{wb}
\end{array}$
Hence, the correct answer is option: 'A'.