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

$q=15\times10^{-6}C$

The velocity is v=1050 m/s

Angle is

$\theta=11.5^o$

To find

a. Magnetic field inside the solenoid

b. The force

Explanation

a. The magnetic field is

$B=\mu_onI$

n is the number of turns per unit length.

Thus,

$\begin{gathered} n=\frac{N}{l} \\ \Rightarrow n=\frac{200}{0.2}=1000 \end{gathered}$

So,

$B=4\pi\times10^{-7}\times1000\times3.25=4.08\times10^{-3}=4.08\text{ mT}$

b. The magnetic force is given by:

$\begin{gathered} F=\text{qvBsin}\theta \\ \Rightarrow F=15\times10^{-6}\times1050\times4.08\times10^{-3}\sin 11.5=1.28\times10^{-5}N \end{gathered}$

Conclusion

a.The magnetic field is 4.08 mT

b.The magnetic force is

$1.28\times10^{-5}N$