Question

A 200-turn solenoid is 20.0 cm long and carries a current of 3.25 A. 1. Find the force in [µN] exerted on a 15.0x10-6 C charged particle moving at 1050 m/s through the interior of the solenoid, at an angle of 11.5° relative to the solenoid’s axis. Find the magnetic field inside the solenoid in [mT]. = 4.08 mT

Answer 1

Given:

• Number ot turns, N = 200

,

• Length = 20.0 cm ==> 0.2 m

,

• Current = 3.25 A

,

• Charge, q = 15.0 x 10⁻⁶ C

,

• Speed = 1050 m/s

,

• Angle = 11.5 degrees

Let's solve for the following:

• (a), The force exerted on the particle.

To find the force exerted on the particle, apply the formula:

`\begin{gathered} F=qV* B \\ \\ F=\text{qVBsin}\theta \end{gathered}`

Where:

B is the magnetic field strength. = 4.08 mT

Thus, we have:

`\begin{gathered} F=(15*10^(-6))*1050*(4.08*10^(-3))*\sin 11.5 \\ \\ F=1.28*10^(-5)N \end{gathered}`

Therefore, the force exterted on the charged particle is 1.28 x 10⁻⁵ N.

ANSWER:

1.28 x 10⁻⁵ N