Q141 / 180 · COMEDK 2013 · Physics
A vertical circular coil of radius $0.1 \mathrm{~m}$ and having $10$ turns carries a steady current. When the plane of the coil is normal to the magnetic meridian, a neutral point is observed at the centre of the coil. If $B_{H}=0.314 \times 10^{-4} \mathrm{~T}$, the current in the coil is
- Correct answerA$0.5 \mathrm{~A}$
- B$0.25 \mathrm{~A}$
- C$2 \mathrm{~A}$
- D$1 \mathrm{~A}$
Solution
Given, $R=0.1 \mathrm{~m}, N=10, B_{H}=0.314 \times 10^{-4} \mathrm{~T}$
For neutral point, the magnetic field at the centre of coil carrying current is equal to the horizontal component of earth's field.
$$
\text { i.e. } \quad \begin{aligned}
B_{H} &=\frac{\mu_{0} N I}{2 R} \\
\Rightarrow \quad I &=\frac{2 B_{H} R}{\mu_{0} N} \\
&=\frac{2 \times 0.314 \times 10^{-4} \times 0.1}{4 \pi \times 10^{-7} \times 10} \\
&=\frac{0.314 \times 10^{-4}}{2 \times 3.14 \times 10^{-7} \times 100}=05 \mathrm{~A}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.