Question:medium

A steel wire of length \(l\) and magnetic moment \(M\) is bent into a semicircular arc of radius \(R\). The new magnetic moment is

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When a magnetic wire is bent into a semicircle, the pole separation becomes the diameter: \[ 2R=\frac{2l}{\pi} \] Magnetic moment depends on pole strength \(\times\) pole separation.
Updated On: Jun 22, 2026
  • \(M\)
  • \(\dfrac{2RM}{\pi l}\)
  • \(\dfrac{2M}{\pi}\)
  • \(\dfrac{2\pi RM}{l}\)
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The Correct Option is C

Solution and Explanation

Step 1: Express the original magnetic moment.
A straight magnetised wire of length $l$ and pole strength $m$ has magnetic moment \[ M = m\, l \] so the pole strength is $m = \dfrac{M}{l}$.
Step 2: Note that pole strength stays the same.
Bending the wire does not change the pole strength $m$; only the separation between the poles changes.
Step 3: Relate the wire length to the arc radius.
When the wire becomes a semicircle of radius $R$, its length equals the arc length, so \[ l = \pi R \quad\Rightarrow\quad R = \frac{l}{\pi} \]
Step 4: Find the new pole separation.
The two ends of a semicircle are separated by the diameter, so the new effective magnetic length is \[ 2R = \frac{2l}{\pi} \]
Step 5: Compute the new magnetic moment.
The new moment is pole strength times new length, \[ M' = m \times 2R = \frac{M}{l} \times \frac{2l}{\pi} \]
Step 6: Simplify and conclude.
The $l$ cancels, leaving \[ M' = \frac{2M}{\pi} \] so the new magnetic moment is \[ \boxed{\dfrac{2M}{\pi}} \]
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