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A semi-circular loop of radius $30 \mathrm{~cm}$ wire carries current $6 \mathrm{~A}$. An uniform magnetic field $0.5 \mathrm{~T}$ is present perpendicular to the plane of the loop. What is the magnitude of force exerted on the wire?
Options:
Solution:
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Verified Answer
The correct answer is:
$1.8 \mathrm{~N}$
Given, the radius of a circular loop,
$$
R=30 \times 10^{-2} \mathrm{~m}
$$
current, $I=6 \mathrm{~A}$ and magnetic field, $B=0.5 \mathrm{~T}$.
Force exerted on the wire,
$$
\begin{array}{rlrl}
& F & =I L B \sin \theta \\
\Rightarrow \quad & F & =6 \times 2 \times 30 \times 10^{-2} \times 0.5 \times \sin 90^{\circ} \\
\Rightarrow \quad & & {\left[\because \theta=90^{\circ} \text { and } \sin 90^{\circ}=1\right]} \\
& =1.8 \mathrm{~N}
\end{array}
$$
Hence, the correct option is (2).
$$
R=30 \times 10^{-2} \mathrm{~m}
$$
current, $I=6 \mathrm{~A}$ and magnetic field, $B=0.5 \mathrm{~T}$.
Force exerted on the wire,
$$
\begin{array}{rlrl}
& F & =I L B \sin \theta \\
\Rightarrow \quad & F & =6 \times 2 \times 30 \times 10^{-2} \times 0.5 \times \sin 90^{\circ} \\
\Rightarrow \quad & & {\left[\because \theta=90^{\circ} \text { and } \sin 90^{\circ}=1\right]} \\
& =1.8 \mathrm{~N}
\end{array}
$$
Hence, the correct option is (2).
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