A wire of length 6.28 m is bent into a circular coil of 2 turns. If a current of 0.5 A exists in the coil, the magnetic moment of the coil is, in ~A-m2

Question

A wire of length 6.28 m is bent into a circular coil of 2 turns. If a current of 0.5 A exists in the coil, the magnetic moment of the coil is, in ~A-m2  (A)  (π /4)  (B)  (1/4)  (C)  π  (D) 4 π

Tardigrade Admin · Accepted Answer

Given, N = 2 i=0.5A When a wire is bent into a circular coil of 2 turns then 4π r=6.28 r=(6.28/4× π ) =(6.28/4× 3.14) r=(1/2) Area, A=π r2=(π /4) Magnetic moment = NiA =2× 0.5× (π /4) =(π /4)

Q. A wire of length 6.28 m is bent into a circular coil of 2 turns. If a current of 0.5 A exists in the coil, the magnetic moment of the coil is, in $A - m^{2}$

$\frac{π}{4}$

$\frac{1}{4}$

$π$

4 $π$

Given, N = 2 $i = 0.5 A$ When a wire is bent into a circular coil of 2 turns then
$4 π r = 6.28$ $r = \frac{6.28}{4 \times π}$ $= \frac{6.28}{4 \times 3.14}$ $r = \frac{1}{2}$ Area, $A = π r^{2} = \frac{π}{4}$ Magnetic moment = NiA $= 2 \times 0.5 \times \frac{π}{4}$ $= \frac{π}{4}$

Q. A wire of length 6.28 m is bent into a circular coil of 2 turns. If a current of 0.5 A exists in the coil, the magnetic moment of the coil is, in A−m2

4π​

41​

π

4 π

Given, N = 2 i=0.5A When a wire is bent into a circular coil of 2 turns then 4πr=6.28 r=4×π6.28​ =4×3.146.28​ r=21​ Area, A=πr2=4π​ Magnetic moment = NiA =2×0.5×4π​ =4π​

Q. A wire of length 6.28 m is bent into a circular coil of 2 turns. If a current of 0.5 A exists in the coil, the magnetic moment of the coil is, in $A - m^{2}$

$\frac{π}{4}$

$\frac{1}{4}$

$π$

4 $π$

Given, N = 2 $i = 0.5 A$ When a wire is bent into a circular coil of 2 turns then
$4 π r = 6.28$ $r = \frac{6.28}{4 \times π}$ $= \frac{6.28}{4 \times 3.14}$ $r = \frac{1}{2}$ Area, $A = π r^{2} = \frac{π}{4}$ Magnetic moment = NiA $= 2 \times 0.5 \times \frac{π}{4}$ $= \frac{π}{4}$