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IRINA_888 [86]
3 years ago
16

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 t

he magnetic force is 7.5 × 10–2 N, what is the magnitude of the charge?
Physics
2 answers:
nekit [7.7K]3 years ago
4 0

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

artcher [175]3 years ago
3 0

Answer:

answer is C on edg

Explanation:

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