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Q)

What is the magnitude of magnetic force per unit length on a wire carrying a current $8\;A$ and making an angle of $30^{\circ}$ with the direction of a uniform magnetic field of $0.15\;T$?

$\begin{array}{1 1} 0.6 \;Nm \\ 6\;Nm \\ 0.6\;N/m \\ 0.6\;Nm \end{array} $

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A)
Solution :
$I=8\;A$
$\theta =30^{\circ}$
$B= 0.15\;T$
$l=1m$
$F= I(l \times B)$
$\quad= IlB \sin \theta$
$\quad= 8 \times 1 \times 0.15 \times \sin 30^{\circ}$
$\quad= 8 \times 1 \times 0.15 \times \large\frac{1}{2}$
$\quad= 0.6 N/m$
Answer : $0.6\;N/m $
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