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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 $B_{H}=0.314 \times 10^{-4} \mathrm{~T}$, then the current in the coil is
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Verified Answer
The correct answer is:
$0.5 \mathrm{~A}$
Given, radius of the coil $=0.1 \mathrm{~m}$
No. of turns in the coil $=10$
Horizontal component of magnetic field,
$$
B_{H}=0.314 \times 10^{-4} \mathrm{~T}
$$
The magnetic field at the centre of current carrying coil is given by the formula, $B=\frac{\mu_{0} n I}{2 R}$
According to question, magnetic field due to the coil obtains neutral point, in earth's magnetic field.
Hence, horizontal component of magnetic field,
$$
\begin{aligned}
& B_{H}=\frac{\mu_{0} n I}{2 R} \\
\text { Thus, } \quad I &=\frac{2 R B_{H}}{\mu_{0} n} \\
\Rightarrow \quad I &=\frac{2(0.1)\left(0.314 \times 10^{-4}\right)}{10 \times 4 \times(3.142) \times 10^{-7}}=0.5 \mathrm{~A}
\end{aligned}
$$
No. of turns in the coil $=10$
Horizontal component of magnetic field,
$$
B_{H}=0.314 \times 10^{-4} \mathrm{~T}
$$
The magnetic field at the centre of current carrying coil is given by the formula, $B=\frac{\mu_{0} n I}{2 R}$
According to question, magnetic field due to the coil obtains neutral point, in earth's magnetic field.
Hence, horizontal component of magnetic field,
$$
\begin{aligned}
& B_{H}=\frac{\mu_{0} n I}{2 R} \\
\text { Thus, } \quad I &=\frac{2 R B_{H}}{\mu_{0} n} \\
\Rightarrow \quad I &=\frac{2(0.1)\left(0.314 \times 10^{-4}\right)}{10 \times 4 \times(3.142) \times 10^{-7}}=0.5 \mathrm{~A}
\end{aligned}
$$
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