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A current of 10 ampere is flowing in a wire of length $1.5 \mathrm{~m}$. A force of $15 \mathrm{~N}$ acts on it when it is placed in a uniform magnetic field of 2 tesla. The angle between the magnetic field and the direction of the current is
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The correct answer is:
$30^{\circ}$
Force on a current carrying wire of length \(l\) and current i, placed in uniform magnetic field \(B\) is given by
\(\begin{aligned}
& \mathrm{F}=\mathrm{i}(\overrightarrow{\mathrm{l}} \times \overrightarrow{\mathrm{B}}) \\
& \mathrm{F}=\mathrm{IlB} \sin \theta \text { or } \sin \theta=\frac{\mathrm{F}}{\mathrm{IlB}}
\end{aligned}\)
\(\sin \theta=\frac{15}{10 \times 1.5 \times 2}=\frac{1}{2} \text { or } \theta=30^{\circ}\)
\(\begin{aligned}
& \mathrm{F}=\mathrm{i}(\overrightarrow{\mathrm{l}} \times \overrightarrow{\mathrm{B}}) \\
& \mathrm{F}=\mathrm{IlB} \sin \theta \text { or } \sin \theta=\frac{\mathrm{F}}{\mathrm{IlB}}
\end{aligned}\)
\(\sin \theta=\frac{15}{10 \times 1.5 \times 2}=\frac{1}{2} \text { or } \theta=30^{\circ}\)
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