From the above figure, it is clear that the neutral point will lie on equitorial plane. B0=4πμ0mr21BH=2B0cosθ⇒0.4×10−4=4πr22μ0m⋅r7×10−2(∵BH=0.4G=0.4×10−4T)[∵cosθ=r7 and r=(72+182)1/2]
⇒0.4×10−4=2×10−7×(72+182)3/2m×7×104
⇒m=144×10−2×(373)3/2...(i) ∵ Magnetic moment, M=m×2l=m×10014...(ii) From Eqs. (i) & (ii), we get M=144×10−2×(373)3/2×10014=4×10−4×7203.82=2.88J/T
