Question

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.

Answer 1

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.