Question

A long wire is bent into a circular coil of one turn having radius ‘R’ and a current T is passed through it 

1. Name the law to find the direction of magnetic field due to this current loop

2. Find an expression for magnetic field produced by this current loop at its centre

3. If the same wire is bent to a smaller radius Y having ‘n’ turns and send same current through it. Find the ratio of magnetic field at the centre in two cases.

Answer 1

1. Right hand screw rule.

2. The magnetic field at a distance × from centre of loop is given by

B = \(\frac{\mu_0R^2I}{2(x^2+R^2)^{3/2}}\)

At the Center (x = 0)  B = \(\frac{\mu_0I}{2R}\)

3. B₁ = \(\frac{\mu_0I}{2R}\)

If wire is bend into smaller radius of n turns.

\(B_2=\frac{\mu_0nI}{R_2}\)  But R₂ = R/n

\(B_2= \frac{\mu_0nI}{\frac{R}{n}}\)     B₂ = \(n^2\frac {\mu_0nI}{R}\)

B₂ = n² B₁