A straight wire carrying current I is turned into a circular loop. If the magnitude of magnetic moment associated with it in M.K.S unit is M, then length of the wire will be:

Question

A straight wire carrying current I is turned into a circular loop. If the magnitude of magnetic moment associated with it in M.K.S unit is M, then length of the wire will be: (A) 4 π IM (B) √4 π IM  (C) √(4 π M / I ) (D) ( M π/4 I )

Tardigrade Admin · Accepted Answer

Magnetic moment of circular loop carrying current M=I A=I(π R2) M=I π((L/2 π))2 ⇒ (I L2 π/4 π2) ⇒ (Π2/4 π)=M L =√(4 π M / I )

Q. A straight wire carrying current I is turned into a circular loop. If the magnitude of magnetic moment associated with it in M.K.S unit is $M$ , then length of the wire will be:

$4 π I M$

$4 π I M$

$\frac{4 π M}{I}$

$\frac{M π}{4 I}$

Magnetic moment of circular loop carrying current
$M = I A = I (π R^{2})$
$M = I π (\frac{L}{2 π})^{2}$
$\Rightarrow \frac{I L ^{2} π}{4 π ^{2}}$
$\Rightarrow \frac{Π ^{2}}{4 π} = M$
$L = \frac{4 π M}{I}$

Q. A straight wire carrying current I is turned into a circular loop. If the magnitude of magnetic moment associated with it in M.K.S unit is M, then length of the wire will be:

4πIM

4πIM​

I4πM​​

4IMπ​

Magnetic moment of circular loop carrying current M=IA=I(πR2) M=Iπ(2πL​)2 ⇒4π2IL2π​ ⇒4πΠ2​=M L=I4πM​​

Q. A straight wire carrying current I is turned into a circular loop. If the magnitude of magnetic moment associated with it in M.K.S unit is $M$ , then length of the wire will be:

$4 π I M$

$4 π I M$

$\frac{4 π M}{I}$

$\frac{M π}{4 I}$

Magnetic moment of circular loop carrying current
$M = I A = I (π R^{2})$
$M = I π (\frac{L}{2 π})^{2}$
$\Rightarrow \frac{I L ^{2} π}{4 π ^{2}}$
$\Rightarrow \frac{Π ^{2}}{4 π} = M$
$L = \frac{4 π M}{I}$