Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The length of a magnetized iron bar is $L$ and its magnetic moment is $M$ . When this bar is bent to form a semicircle its magnetic moment is:

$M$

$\frac{M\pi}{2}$

$\frac{M}{2\pi}$

$\frac{2M}{\pi}$

Initial State: The magnetic moment $M$ of a straight bar magnet is defined as the product of its pole strength ( $m$ ) and the separation distance between the poles ( $L$ ). $M = m \times L$
Bending Geometry: When the bar is bent into a semicircle, the length of the arc remains $L$ . Let the radius of the semicircle be $r$ . From the geometry of a circle: $\text{Arc Length} = \pi r = L \implies r = \frac{L}{\pi}$
New Effective Length: The new magnetic moment ( $M'$ ) depends on the vector displacement (shortest straight-line distance) between the two poles. For a semicircle, this distance is the diameter ( $2r$ ). Reference to displacement vectors can be found in kinematics . $L' = 2r = \frac{2L}{\pi}$
Calculation: $M' = m \times L' = m \times \left(\frac{2L}{\pi}\right) = \frac{2}{\pi} (m L)$ Substituting $M = mL$ : $M' = \frac{2M}{\pi}$

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