JEE Main 17 March 2021 Shift 1 Solved Paper

Given, number of turns per unit length, n=1000
Relative magnetic permeability, µt=500
Current flowing through solenoid, I=5A
Magnetic permeability of free space,
µ0=4π×10−7H/m
We know that, magnetic field inside the solenoid can be given as
B=µnI ....(i)
where, µ= permeability of medium.
As ‌‌µ=µrµ0...(ii)
From Eqs. (i) and (ii), we get
∴‌‌B=µrµ0nl
=500×4π×10−7×1000×5
=100π×10−7×105
=πT