MHTCET 2016 Physics Solved Paper

(c) Given, B0 = magnetic field after insertion of iron rod - 2000 Am−1
Magnetic flux, ϕ = 6 x 10−4 Wb
Area of cross-section, A = 3 cm2 = 3 x 10−4 m
magnetic permeability of the rod,
µr=

B

B0

= ϕA×B0[∵ϕ=BA]
So, µr=

B

2000Am−1

∴ µr=

6×10−4Wb

3×10−4m2

×

1

2000Am−1

= 10−3WbA−1m−1