A solenoid of 300 turns/m is carrying a current of 5A The length of the solenoid is 0 5m and - Physics - Moving Charges And Magnetism

Question

A solenoid of 300 turns/m is carrying a current of 5A The length
of the solenoid is 0 5m and has a radius of 1cm Find the magnitude
of magnetic field inside the solenoid

Swati · Accepted Answer

The magnitude of magnetic field inside the solenoid is given
as,

B=μonIwhere,μo=4π×10-7 Tesla/Amp
mn=number of turns per unit lengthI=current
On substituting the values we get,

B=4π×10-7 Tesla/Amp
m×300 turns/m×0 5 AmpB=1
88×10-4 Tesla

A solenoid of 300 turns/m is carrying a current of 5A.The length of the solenoid is 0.5m and has a radius of 1cm.Find the magnitude of magnetic field inside the solenoid.