Question

A long solenoid has 103 turns/cm and carries current i. An electron moves within the solenoid in a circle of radius 2.60 cm perpendicular to the solenoid axis. The speed of the electron is 1.38 × 107 m/s. Find the current i in the solenoid.

Answer 1

Answer:

0.23348 A

Explanation:

B = Magnetic field

v = Velocity of electron = $1.38\times 10^7\ m/s$

q = Charge of electron = $1.6\times 10^{-19}\ C$

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

r = Radius of circle = 0.026 m

N = Number of turns = 103 turns/cm = $103\times 100\ turns/m$

I = Current

The magnetic and centripetal force will be balanced

$Bqv=m\frac{v^2}{r}\\\Rightarrow B=\frac{mv}{qr}$

The magnetic field in solenoid is given by

$B=N\mu_0 I\\\Rightarrow I=\frac{B}{N\mu_0}$

From the first equation

$I=\frac{\frac{mv}{qr}}{N\mu_0}\\\Rightarrow I=\frac{mv}{N\mu_0qr}\\\Rightarrow I=\frac{9.11\times 10^{-31}\times 1.38\times 10^7}{103\times 100\times 4\pi \times 10^{-7}\times 1.6\times 10^{-19}\times 0.026}\\\Rightarrow I=0.23348\ A$

The current in the solenoid is 0.23348 A