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?

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

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

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