Question

A charged particle moves at 2. 5 × 104 m/s at an angle of 25° to a magnetic field that has a field strength of 8. 1 × 10–2 T. If the magnetic force is 7. 5 × 10–2 N, what is the magnitude of the charge? 3. 7 × 10–5 C 4. 1 × 10–5 C 8. 8 × 10–5 C 1. 0 × 10–4 C.

Answer 1

The force acting on a moving charge is known as the magnetic force. The force acting on the charge will be 3.75 N.

What is the force exerted on the charge?

Magnetic fields only exert a force on a moving electric charge. A moving charge generates a magnetic field. With an increase in charge and magnetic field strength, this force rises.

when charges have higher velocities, the force is stronger. However, the magnetic force is always perpendicular to the velocity.

Mathematically the force exerted on the charge will be

F=qvBsinα

F= force acting on the charge= 7.5 ×10⁻²

v = velocity of charge=2.5 ×10⁴

q = charge=?

B= 8.1 ×10⁻²

F=qvBsinα

$\rm{q= \frac{F}{vBsina}}$

$\rm{q= \frac{7.5\times 10^-2}{2.4\times 10^-2 \times 8.1\times 10^-2 sin25^0}}$

q = $8.8 \times 10^ {-5}\rm C$

Hence the value of the charge will be $8.8 \times 10^ {-5}\rm C$.

To learn more about the force acting on charge refer to ;

brainly.com/question/451411