A particle of charge perpendicular to a magnetic field of magnitude 0.01T if the charge is 5c calculate the magnetic force exerted on the particle.

The magnetic force $F$ on a charged particle moving in a magnetic field is given by the formula:

$$F = QvB \sin\theta$$

where:

• $Q$ is the magnitude of the charge.
• $v$ is the speed of the particle.
• $B$ is the magnetic field strength.
• $\theta$ is the angle between the velocity vector and the magnetic field vector.

From the problem statement and the provided information:

• Charge $Q = 5\, \text{C}$
• Magnetic field strength $B = 0.01\, \text{T}$
• Velocity $v = 1.5\, \text{m/s}$
• Angle $\theta = 90^\circ$ (since the particle is moving perpendicularly to the magnetic field)

Substitute the given values into the magnetic force formula:

$$F = (5\, \text{C})(1.5\, \text{m/s})(0.01\, \text{T}) \sin(90^\circ)$$

The sine of $90^\circ$ is 1:

$$\sin(90^\circ) = 1$$

$$F = (5)(1.5)(0.01)(1)\, \text{N}$$ $$F = (7.5)(0.01)\, \text{N}$$ $$F = 0.075\, \text{N}$$

The magnetic force exerted on the particle is $\text{0.075 N}$.