Apositive charge of 0.026 c moves horizontally to the right at a speed of 443.592 m/s and enters a magnetic field directed vertically downward. if it experiences a force of 22.182 n, what is the magnetic field strength ?

Magnetic field, B = 1.9232 T

Explanation:

Given data:

Value of the charge, Q = 0.026 C

Speed, V = 443.592 m/s

Force experienced, F = 22.182

Now,

the Force (F) experienced by a charge in a magnetic field is given as:

F = QVBsinθ

where,

B is the magnetic field

Angle between the magnetic field and the velocity.

since, the velocity is in horizontal direction and the magnetic field is downwards. Therefore, the angle θ = 90°

thus, we have

22.182 = 0.026 × 443.592 × B × sin90°

or

B = 1.9232 T

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