Question

A charged particle moving through a magnetic field at right angles to the field with a speed of 24.7 m/s experiences a magnetic force of 2.38 10-4 N. Determine the magnetic force on an identical particle when it travels through the same magnetic field with a speed of 5.64 m/s at an angle of 21.2° relative to the magnetic field. N

Answer 1

Answer:

Explanation:

let the charge is q. velocity, v = 24.7 m/s

magnetic force, F = 2.38 x 10^-4 N

Let the magnetic field is B.

Velocity, v' = 5.64 m/s

angle, θ = 21.2°

The force experienced by a charged particle placed in a magnetic field is given by

F = q x v x B x Sinθ

in first case

2.38 x 10^-4 = q x 24.7 x B x Sin 90 .... (1)

in second case

F = q x 5.64 x B x Sin 21.2°      .... (2)

Divide equation (2) by equation (1), we get

$\frac{F}{2.38\times 10^{-4}}=\frac{5.64\times Sin 21.2}{24.7\times Sin 90}$

F = 1.97 x 10^-5 N