Given, cross-sectional area of iron bar, A=2×10−5m2 Magnetising field, H=2400A∕m Magnetic flux, φ=2.4π×10−5‌Wb We know that, φ=BA ⇒‌‌B=‌
φ
A
=‌
2.4π×10−5
2×10−5
=1.2πT ∴ Again, B=µH
⇒‌‌µ=‌
˙B
H
=‌
1.2Ï€
2400
⇒µ=5π×10−4
Again, we know that, magnetic susceptibility, χ‌=µr−1=‌