First slide

Force on a current carrying wire placed in a magnetic field

Question

A current of 10 amp is flowing in a wire of length 1 .5 metre. A force of 15 newton 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 current is

Easy

A

30°
B

45°
C

60°
D

90°

Solution

The magnitude of force is given by
$\begin{array}{l} F = | i (l \times B) | = ilBsin θ \\ 15 = 10 \times 1 .5 \times 2 \sin θ or \sin θ = 1 / 2 \\ θ = \sin^{- 1} (1 / 2) = 30^{\circ} \end{array}$

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