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  4. A current of 10 A is flowing in a wire of length 1.5 m. A force of 15 N acts on it when it is placed in a uniform magnetic field of 2 T. The angle between the magnetic field and the direction of the current is

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Q. $A$ current of $10 \,A$ is flowing in a wire of length $1.5\, m$. A force of $15\, N$ acts on it when it is placed in a uniform magnetic field of $2\, T$. The angle between the magnetic field and the direction of the current is

Moving Charges and Magnetism

Solution:

$F=IlB\, sin\,\theta$ or $sin\,\theta = \frac{F}{IlB}$
$sin\,\theta = \frac{15}{10\times1.5\times2}=\frac{1}{2}$
or $\theta=30^{\circ}$