Question

A charged particle with charge +7.5 nC moves to the right with a velocity of 25 m/s in a magnetic field of magnitude 7 mT which also points to the left. What is the magnitude of the force on the charged particle?

Answer 1

0N

Step-by-step explanation:

According to Lorentz, the magnitude of the magnetic force, F, on a charged particle of magnitude, Q, moving at a speed, v, in a magnetic field of intensity, B, is given by;

F = Q x v x B x sinθ ---------------(i)

Where;

θ = angle between the directions of the velocity, v, and magnetic field, B

Now, from the question;

Q = 7.5nC = 7.5 x 10⁻⁹C

v = 25m/s

B = 7mT = 7 x 10⁻³ T

θ = 180° (since the particle moves to the right in a magnetic field which points to the left)

Substitute these values into equation (i) as follows;

F = 7.5 x 10⁻⁹ x 25 x 7 x 10⁻³ x sin(180)

F = 7.5 x 10⁻⁹ x 25 x 7 x 10⁻³ x 0

F = 0

Therefore, the magnitude of the force on the charged particle is 0N

Answer 2

The magnitude of the force on the charged particle is 1.3125nN

Step-by-step explanation:

Given;

magnitude of charge, q = +7.5 nC

velocity of the charged particle, v = 25 m/s

strength of magnetic field, B = 7 mT

Magnitude of force on charged particle in magnetic field is given as;

F = qvB

where;

F is the force on the moving particle

q is the charged particle

B is magnetic field strength

Substitute in the given values in the above equation, we will have;

F = 7.5 x 10⁻⁹ x 25 x 7 x 10⁻³

F = 1.3125 x 10⁻⁹ N

F = 1.3125nN

Therefore, the magnitude of the force on the charged particle is 1.3125nN