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.
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 NEET…
NEET UG 2013 — Physics Electromagnetism
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 
Held on 30 Apr 2013 · Verified 9 Jul 2026.
Options
- A
M
- B
π3M
- C
π2M
- D
2M
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
NEET UG Electromagnetism in other years
More NEET UG Electromagnetism questions
Two point charges +2μC and -2μC are placed 10 cm apart. The electric field at the midpoint is:
A model for quantized motion of an electron in a uniform magnetic field $B$ states that the flux passing through the orbit of the electron is $n(h / e)$ where $n$ is an integer, $h$ is Planck's constant and $e$ is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be ( $m$ is the mass of the electron)
A 2 amp current is flowing through two different small circular copper coils having radii ratio $1: 2$. The ratio of their respective magnetic moments will be
Two identical charged conducting spheres $A$ and $B$ have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is $F$. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres $A$ and $B$ (Radii of $A$ and $B$ are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as :
An electron (mass $9 \times 10^{-31} \mathrm{~kg}$ and charge $1.6 \times 10^{-19} \mathrm{C}$ ) moving with speed $\mathrm{c} / 100(\mathrm{c}=$ speed of light) is injected into a magnetic field $\vec{B}$ of magnitude $9 \times 10^{-4} \mathrm{~T}$ perpendicular to its direction of motion. We wish to apply an uniform electric field $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light $\mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}$ )
Other Physics topics for NEET UG
Keep practicing Electromagnetism
Work through every NEET UG Electromagnetism PYQ, year by year.