JEE Main 16 March 2021 Shift 1 Solved Paper

The given situation can be shown as belowFrom the above figure, it is clear that the neutral point will lie on equitorial plane.
B0=‌

µ0m

4π

‌

1

r2

BH=2B0‌cos‌θ
⇒‌‌0.4×10−4=‌

2µ0m

4πr2

⋅‌

7×10−2

r

 (∵BH=0.4G=0.4×10−4T)
[∵cos‌θ=‌

7

r

and r=(72+182)1∕2]
⇒‌‌m=‌

4×10−2×(373)3/2

14

...(i)
∵ Magnetic moment, M=m×2l=m×‌

14

100

...(ii)
From Eqs. (i) & (ii), we get
M‌‌=‌

4×10−2×(373)3/2

14

×‌

14

100

‌‌=4×10−4×7203.82=2.88J/T