A charged particle moves at 2.5 * 104 m/s at an angle of 25° to a magnetic field that has a field strength of 8.1 * 10-2 T. If the magnetic force is 7.5 * 10-2 N, what is the magnitude of the charge?

Question

Lesly Barajas

Physics

30 June, 04:10

A charged particle moves at 2.5 * 104 m/s at an angle of 25° to a magnetic field that has a field strength of 8.1 * 10-2 T. If the magnetic force is 7.5 * 10-2 N, what is the magnitude of the charge?

+1

Answers (2)

Know the Answer?

Miley Foster · Answer 1

answer is C on edg

Andrew Cobb · Answer 2

the magnitude of charge=q=8.76 x 10⁻⁵C

Explanation:

the magnetic force Fm is given by

Fm = q V B sinθ

q = charge

v = velocity = 2.5 x 10⁴ m/s

B = magnetic field strength = 8.1 x 10⁻²T

Fm = magnetic force = 7.5 x 10⁻² N

θ=25°

so 7.5 x 10⁻² = q (2.5 x 10⁴) (8.1 x 10⁻²) sin25

q=8.76 x 10⁻⁵C