Question : Two wires of the same length and shaped into a square of side 'a' and a circle with radius 'r'. If they carry same current, the ratio of their magnetic moment is
(a) 2:π
(b) π:2
(c) π:4
(d) 4:π
Doubt by Lakshita
Solution :
Let L be the length of the wire.
Case I : When the wire is bent in the shape of the square.
L=Perimeter of square
L=4a (where a is the side of square)
a=L/4 — (1)
Now, Magnetic Moment of Square Loop
M1=IA1
M1=I(a²)
M1=I(L/4)² [From Equation (1)]
M1=IL²/16 — (2)
Case II : When the wire is bent in the shape of the circle.
L=Perimeter of square
L=4a (where a is the side of square)
a=L/4 — (1)
Now, Magnetic Moment of Square Loop
M1=IA1
M1=I(a²)
M1=I(L/4)² [From Equation (1)]
M1=IL²/16 — (2)
Case II : When the wire is bent in the shape of the circle.
L=Perimeter of the circle
L=2πr r=L/2π — (3)
Now, Magnetic Moment of Circular Loop
M2=IA2 M2=I(πr²)
L=2πr r=L/2π — (3)
Now, Magnetic Moment of Circular Loop
M2=IA2 M2=I(πr²)
M2=I[π(L/2π)²]
M2=I[π(L²/4π²)]
M2=IL²/4π — (4)
M2=IL²/4π — (4)
Dividing equation (1) by (2)
M1/M2 = [IL²/16]/[IL²/4π]
M1/M2 = 4π/16
M1/M2 = π/4
M1:M2=π:4
M1/M2 = 4π/16
M1/M2 = π/4
M1:M2=π:4
Hence, (c) π:4 is the correct option.
