Two wires of same length are shaped into a square and a circle. If they carry same current, ratio of the magnetic moments isa)2 : andpi;b)andpi; : 2c)andpi; : 4d)4 : andpi;Correct answer is option 'C'. Can you explain this answer?

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Two wires of same length are shaped into a square and a circle. If they carry same current, ratio of the magnetic moments is

a)
2 : π
b)
π : 2
c)
π : 4
d)
4 : π

Correct answer is option 'C'. Can you explain this answer?

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Two wires of same length are shaped into a square and a circle. If the...

In the case everything is same except area.,so we will consider the area. Also emf is directly proportional to area by formula(emf=BA/T) now ratio of areas is pie :4

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Two wires of same length are shaped into a square and a circle. If the...

To find the ratio of the magnetic moments, we need to compare the magnetic fields generated by the square and the circle.

For a square loop carrying current I, the magnetic field at the center of the loop can be found using Ampere's law:

B_square = (μ0 * I) / (2 * a)

where μ0 is the permeability of free space and a is the side length of the square.

For a circular loop carrying current I, the magnetic field at the center of the loop can be found using the Biot-Savart law:

B_circle = (μ0 * I * R) / (2 * R^2)

where R is the radius of the circle.

Since the wires have the same length, the perimeter of the square is equal to the circumference of the circle:

4a = 2πR

R = (2a) / π

Substituting this value of R into the expression for B_circle, we get:

B_circle = (μ0 * I * (2a / π)) / (2 * (2a / π)^2)

B_circle = (μ0 * I * π) / (4a)

Now we can find the ratio of the magnetic fields:

B_square / B_circle = [(μ0 * I) / (2a)] / [(μ0 * I * π) / (4a)]

B_square / B_circle = (4a) / (2a * π)

B_square / B_circle = 2 / π

Therefore, the ratio of the magnetic moments is 2 : π.

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