Answer :
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:
[tex]\begin{gathered} F=qV\times B \\ \\ F=\text{qVBsin}\theta \end{gathered}[/tex]Where:
B is the magnetic field strength. = 4.08 mT
Thus, we have:
[tex]\begin{gathered} F=(15\times10^{-6})\times1050\times(4.08\times10^{-3})\times\sin 11.5 \\ \\ F=1.28\times10^{-5}N \end{gathered}[/tex]Therefore, the force exterted on the charged particle is 1.28 x 10⁻⁵ N.
ANSWER:
1.28 x 10⁻⁵ N