Question

An electron is moving at a speed of 2.20 ✕ 104 m/s in a circular path of radius of 4.3 cm inside a solenoid. The magnetic field of the solenoid is perpendicular to the plane of the electron's path. The solenoid has 25 turns per centimeter.(a) Find the strength of the magnetic fieldinside the solenoid.(b) Find the current in the solenoid.

Answer 1

Answer:

a) 2.90*10^-6 T

b) 0.092A

Explanation:

a) The magnitude of the magnetic field is given by the formula for the calculation of B when it makes an electron moves in a circular motion:

$B=\frac{m_ev}{qR}$

me: mass of the electron = 9.1*10^{-31}kg

q: charge of the electron = 1.6*10^{-19}C

R: radius of the circular path = 4.3cm=0.043m

v: speed of the electron = 2.20*10^4 m/s

By replacing all these values you obtain:

$B=\frac{(9.1*10^{-31}kg)(2.20*10^4 m/s)}{(1.6*10^{-19}C)(0.043m)}=2.90*10^{-6}T=2.9\mu T$

b) The current in the solenoid is given by:

$I=\frac{B}{\mu_0 N}=\frac{2.90*10^{-6}T}{(4\pi*10^{-7}T/A)(25)}=0.092A=92mA$