A circular loop of radius 20 cm carries a current of 10 A. An electron crosses the plane of the loop with a speed of 2.0 × 10^{6} m s^{−1}. The direction of motion makes an angle of 30° with the axis of the circle and passes through its centre. Find the magnitude of the magnetic force on the electron at the instant it crosses the plane.

#### Solution

Given:

Magnitude of current in the loop, I = 10 A

Radius of the loop, r = 20 cm = 20 × 10^{−2} m

Thus, the magnetic field intensity at the centre is given by \[B = \frac{\mu_0 I}{2r}\]

Now,

Velocity of the electron, v = 2 × 10^{6} m/s

Angle between the velocity and the magnetic field intensity, θ = 30°

Thus, the magnetic force on the electron is given by

\[F = evB\sin \theta\]

\[ = 1 . 6 \times {10}^{- 19} \times 2 \times {10}^6 \times \frac{\mu_0 i}{2R}\sin 30^\circ \]

\[ = 1 . 6 \times {10}^{- 19} \times 2 \times {10}^6 \times \frac{4\pi \times {10}^{- 7} \times 10}{2 \times 20 \times {10}^{- 2}} \times \frac{1}{2}\]

\[ = 16\pi \times {10}^{- 19} \] N