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 xx (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.

Explanation

Here `R = 20cm = 20xx10^(-2) m`
`I = 10 A, V = 2xx10^6 m//s`
`theta = 30^@`
Since `F = e (vecV+vecB)`
`=eVB sin Theta`
`=1.6xx10^(-19)xx2xx10^6 (mu_oi)/(2R) sin 30`
`= 1.6xx10^(-19)xx2xx10^6`
`xx(4pixx10^(-7)xx10)/(2xx20xx10^(-2) xx (1)/(2)`
`= 16pixx10(-19) N`