Question

A current carrying circular loop of area A produces a magnetic field B at its centre. Show that the magnetic moment of the loop ia `(2 BA)/mu_0 sqrt(A/pi)`.

Answer 1

The magnetic moment (μ) of a current-carrying circular loop is given by:

μ = IA

The magnetic field at the centre of a circular loop carrying current is given by:

B = `(mu_0I)/(2R)`

The area of the loop is:

A = πR²

⇒ R = `sqrt(A/pi)`

Now, substitute R back into the equation for B:

B = `(mu_0I)/(2sqrt(A/pi))`

This simplifies to:

B = `(mu_0I(sqrtpi))/(2sqrtA)`

Rearranging the equation to solve for I:

I = `(2BsqrtA)/(mu_0sqrt pi)`

The magnetic moment m of the loop is given by:

m = I.A

Substituting the expression for I:

m = `((2Bsqrt(A))/(mu_0sqrt pi))*A`

This simplifies to:

m = `(2BA sqrtA)/(mu_0 sqrtpi)`