Solved: PHYSICS Question for NEET

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A wire of length $L$ meters carrying a current of $I$ ampere is bent in the form of a circle. What is its magnetic moment?

$\frac{IL^2}{4} \text{ A-m}^2$

$\frac{I \pi L^2}{4} \text{ A-m}^2$

$\frac{2IL^2}{\pi} \text{ A-m}^2$

$\frac{IL^2}{4\pi} \text{ A-m}^2$

Geometry Constraint: When a wire of length $L$ is bent into a circular loop of radius $R$ , the circumference of the loop equals the length of the wire. $L = 2\pi R \Rightarrow R = \frac{L}{2\pi}$
Area Calculation: The area $A$ of the circular loop is given by: $A = \pi R^2 = \pi \left( \frac{L}{2\pi} \right)^2 = \pi \frac{L^2}{4\pi^2} = \frac{L^2}{4\pi}$
Magnetic Moment Formula: The magnetic moment $M$ of a current-carrying loop is defined as the product of the current $I$ and the area $A$ (for a single turn, $N=1$ ). $M = I \times A$
Substitution: Substituting the expression for area: $M = I \times \frac{L^2}{4\pi} = \frac{IL^2}{4\pi}$ The unit is Ampere-meter squared (A-m $^2$ ).

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