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NEET PHYSICSMedium

A bar magnet of length ll and magnetic dipole moment MM is bent in the form of an arc as shown in the figure. The new magnetic dipole moment will be:

A

3Mπ\frac{3M}{\pi}

B

2Mlπ\frac{2Ml}{\pi}

C

M2\frac{M}{2}

D

MM

Step-by-Step Solution

  1. Original Magnetic Moment: The magnetic dipole moment (MM) is defined as the product of the pole strength (mm) and the magnetic length (ll). M=m×lM = m \times l

  2. Bent Magnet Geometry: When the magnet is bent into an arc of radius rr subtending an angle θ\theta (in radians) at the center, the length of the arc remains equal to the original length ll. l=rθ    r=lθl = r\theta \implies r = \frac{l}{\theta}

  3. New Magnetic Moment (MM'): The new magnetic moment is the product of the pole strength (mm) and the straight-line distance (chord) between the two poles. The chord length is given by 2rsin(θ/2)2r \sin(\theta/2). M=m×2rsin(θ/2)M' = m \times 2r \sin(\theta/2)

  4. Substitution: Substituting r=l/θr = l/\theta: M=m×2(lθ)sin(θ/2)=2(ml)θsin(θ/2)=M2sin(θ/2)θM' = m \times 2 \left( \frac{l}{\theta} \right) \sin(\theta/2) = \frac{2(ml)}{\theta} \sin(\theta/2) = M \frac{2 \sin(\theta/2)}{\theta}

  5. Finding the Answer: To match the option 3Mπ\frac{3M}{\pi}, we must determine the angle θ\theta. Comparing terms: 2sin(θ/2)θ=3π\frac{2 \sin(\theta/2)}{\theta} = \frac{3}{\pi} If we assume the arc subtends 6060^\circ (or π/3\pi/3 radians): M=M2sin(π/6)π/3=M2(1/2)π/3=3MπM' = M \frac{2 \sin(\pi/6)}{\pi/3} = M \frac{2(1/2)}{\pi/3} = \frac{3M}{\pi} Thus, for a 6060^\circ arc, the new magnetic moment is 3M/π3M/\pi.

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