Given, θ=60∘=3πrad The new magnetic length of the dipole will be, I′=r Now, M=m⋅I and M′=m⋅I′ ∴M′M=I′I…(1) Now, angle in radians = radius arc length ∴3π=r1⇒1=3πr In the equilateral triangle, marked by dotted lines, I′=r ∴I′1=3π…(2) From and, we get, M′M=3π∴M′=π3M Hence, (C) is the correct answer. Alternative method: Using the short cut formula about bending of a bar magnet, M=θ2Msin(2θ) =3π2Msin30∘=π3M Hence, (C) is the correct answer.
NEET UG 2013 — Physics Electromagnetism
2013mcqeasy
A bar magnet of length l and magnetic dipole moment M is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be 
Official previous-year question
Held on 30 Apr 2013 · Verified 9 Jul 2026.
Options
- A
M
- B
π3M
- C
π2M
- D
2M
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