Question

A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be:

Options

• `"m"/pi`

• `(3"m")/pi`

• `(2"m")/pi`

• `(4"m")/pi`

Answer 1

`bb((4"m")/pi)`

Explanation:

Let a represent the square's area and r represent the circular loop radius.

2πr = 4a ⇒ r = `((2"a")/pi)`

For square

M = (I) a²

For circular loop

M₁ = (I)πr²

⇒ M₁ = (I) (π)`((4"a"^2)/pi^2)`

M₁ = `(4"Ia"^2)/pi`

⇒ M₁ = `(4"M")/pi` (∵ M = Ia²)