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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.

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1 vote

Answer:

0.23348 A

Step-by-step explanation:

B = Magnetic field

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

q = Charge of electron =
1.6* 10^(-19)\ C


\mu_0 = Vacuum permeability =
4\pi * 10^(-7)\ H/m

r = Radius of circle = 0.026 m

N = Number of turns = 103 turns/cm =
103* 100\ turns/m

I = Current

The magnetic and centripetal force will be balanced


Bqv=m(v^2)/(r)\\\Rightarrow B=(mv)/(qr)

The magnetic field in solenoid is given by


B=N\mu_0 I\\\Rightarrow I=(B)/(N\mu_0)

From the first equation


I=((mv)/(qr))/(N\mu_0)\\\Rightarrow I=(mv)/(N\mu_0qr)\\\Rightarrow I=(9.11* 10^(-31)* 1.38* 10^7)/(103* 100* 4\pi * 10^(-7)* 1.6* 10^(-19)* 0.026)\\\Rightarrow I=0.23348\ A

The current in the solenoid is 0.23348 A

User Jparram
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