Question

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 × 10⁶ 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.

Answer 1

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 × 10⁶ 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