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lozanna [386]
3 years ago
14

A particle with a charge of +8.4 C and a speed of 40 m/s enters a uniform magnetic field whose 휇magnitude is 0.33 T. For each of

the three cases in the drawing, calculate the magnitude and direction of the magnetic force on the particle.
Physics
1 answer:
vovikov84 [41]3 years ago
4 0

Explanation:

Given that,

Charge of the particle, q = 8.4 C

Speed of the particle, v = 40 m/s

Magnitude of magnetic field, B = 0.33 T

Let us assumed to find the magnitude and direction of the magnetic force on the particle. Let 30.0 degrees for one, and 150 degrees for two, with a perpendicular charge. So,

F=qvB\ \sin\theta

F_1=8.4\times 40\times 0.33\times \ \sin(30)

F_1=55.44\ N

If angle is 150 degrees

F_2=8.4\times 40\times 0.33\times \ \sin(150)

F_2=55.44\ N

If angle is 90 degrees

F_3=8.4\times 40\times 0.33\times \ \sin(90)

F_3=110.88\ N

Hence, this is the required solution.

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