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

Tardigrade Admin · Accepted Answer

l=2 π r or r=(l/2 π) Area of circular loop A=π r2 Magnetic moment M=i A=i π r2 M=i π × l2 / 4 π2 or 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 MKS unit is M, the length of wire will be

$\frac{4 π}{M}$

$\frac{4 π M}{i}$

$\frac{4 πi}{M}$

$\frac{M π}{i}$

$l = 2 π r$ or $r = \frac{l}{2 π}$
Area of circular loop $A = π r^{2}$
Magnetic moment $M = i A = iπ r^{2}$
$M = iπ \times l^{2} /4 π^{2}$
or $l = \frac{4 π M}{i}$

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

M4π​

i4πM​​

M4πi​​

iMπ​

l=2πr or r=2πl​ Area of circular loop A=πr2 Magnetic moment M=iA=iπr2 M=iπ×l2/4π2 or l=i4πM​​

$\frac{4 π}{M}$

$\frac{4 π M}{i}$

$\frac{4 πi}{M}$

$\frac{M π}{i}$

$l = 2 π r$ or $r = \frac{l}{2 π}$
Area of circular loop $A = π r^{2}$
Magnetic moment $M = i A = iπ r^{2}$
$M = iπ \times l^{2} /4 π^{2}$
or $l = \frac{4 π M}{i}$