What is the magnitude of magnetic force per unit length on a wire carrying a current of $8\,A$ and making an angle of $30^o$ with the direction of a uniform magnetic field of $0.15\, T$ ?.....$Nm^{-1}$
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According to the question, $\mathrm{I}=8 \,\mathrm{A}, \theta=30^{\circ}$ $B=0.15\, \mathrm{T}, \ell=1 \,\mathrm{m}$
The magnitude of magnetic force $F=\mathrm{BI} \ell \sin \theta$
$=0.15 \times 8 \times 1 \times \sin 30^{\circ}=1.2 \times \frac{1}{2}=0.6\, \mathrm{Nm}^{-1}$
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