A wire is wound on a long rod of material of relative permeability μr=4000 to make a solenoid. If the current through the wire is 5 A and number of turns per unit length is 1000 per metre, then the magnetic field inside the solenoid is:
Given that,
Relative permeability, μr=4000
Carried current, I=5 A
Number of turns, n=1000
therefore,
the magnetic intensity,
H=nI
H=1000×5=5000Am−1
Now, the magnetic field inside the solenoid is
B=μH.....(i)
We know that,
μ=μ0μr=4π×10−7×4000
We get,
B=4π×10−7×4000×5000
B=25.12 T