A solenoid 93.0 cm long has a radius of 2.80 cm and a winding of 1100 turns; it

Question

A solenoid 93.0 cm long has a radius of 2.80 cm and a winding of 1100 turns; it carries a current of 3.60 A. Calculate the magnitude of the magnetic field inside the solenoid

A solenoid 1.35 m long and 2.50 cm in diameter carries a current of 22.0 A. The magnetic field inside the solenoid is 25.0 mT. Find the length of the wire forming the solenoid

A long solenoid has 90 turns/cm and carries current i. An electron moves within the solenoid in a circle of radius 2.00 cm perpendicular to the solenoid axis. The speed of the electron is 0.0340c (c = speed of light). Find current i

Explanation / Answer

1)

Magnetic fied due to solenoid is

B=uo*N*I/L =(4pi*10-7)(1100)(3.6)/(0.93)

B=5.35*10-3 T or 5.35 mT

2)

r=d/2 =1.25 cm

Magnetic field of a solenoid is

B=uo*N*I/L

=>N=B*L/uo*I =(25*10-3)*1.35/(4pi*10-7)(22)

N=1221 turns

since

N=Lw/2pir

=>Lw=N*(2pir) =1221*(2pi*1.25*10-2)

Lw=95.9 m

3)

given

v=0.034*(3*108)=1.02*107 m/s

centripetla force =magnetic force

mv2/r=qvB

=>B=mv/qr =(9.11*10-31)(1.02*107)/(1.6*10-19)*0.02

B=2.9*10-3 T

since Magnetic field of a solenoid is

B=uo*n*I

=>I=B/uo*n =(2.9*10-3)/(4pi*10-7)(90*102)

I=0.2568 A

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