Question

A 200-turn solenoid is 20.0 cm long and carries a current of 3.25 A.1. Find the magnetic field inside the solenoid in [mT].2. 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.

Answer 1

Given,

Number of turns, N=200

Length of teh solenoid, l=20 cm=0.2 m

Current, I=3.25 A.

Charge is

`q=15*10^(-6)C`

The velocity is v=1050 m/s

Angle is

`\theta=11.5^o`

To find

a. Magnetic field inside the solenoid

b. The force

Step-by-step explanation

a. The magnetic field is

`B=\mu_onI`

n is the number of turns per unit length.

Thus,

`\begin{gathered} n=(N)/(l) \\ \Rightarrow n=(200)/(0.2)=1000 \end{gathered}`

So,

`B=4\pi*10^(-7)*1000*3.25=4.08*10^(-3)=4.08\text{ mT}`

b. The magnetic force is given by:

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

Conclusion

a.The magnetic field is 4.08 mT

b.The magnetic force is

`1.28*10^(-5)N`