First slide

Magnetic field due to a current carrying circular ring

Question

Two circular coils 1 and 2 are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil. What is the ratio of potential difference applied across them so that the magnetic field at their centres is the same ?

Moderate

A

3
B

4
C

6
D

2

Solution

Magnetic field at the centre of a circular coil is
$B = \frac{μ_{0}}{4 π} \times \frac{2 πi}{r}$
where i is current flowing in the coil and r is radius of coil.
At the centre of coil 1,
$B_{1} = \frac{μ_{0}}{4 π} \times \frac{2 {πi}_{1}}{r_{1}}$ …(i)
At the centre of coil 2,
$B_{2} = \frac{μ_{0}}{4 π} \times \frac{2 {πi}_{2}}{r_{2}}$ …(ii)
But B₁ = B₂
$∴ \frac{μ_{0}}{4 π} \frac{2 {πi}_{1}}{r_{1}} = \frac{μ_{0}}{4 π} \frac{2 {πi}_{2}}{r_{2}} or \frac{i_{1}}{r_{1}} = \frac{i_{2}}{r_{2}}$
As r₁ = 2r₂
$∴ \frac{i_{1}}{2 r_{2}} = \frac{i_{2}}{r_{2}} Or i_{1} = 2 i_{2}$ …(iii)
Now, ratio of potential differences
$\frac{V_{2}}{V_{1}} = \frac{i_{2} \times r_{2}}{i_{1} \times r_{1}} = \frac{i_{2} \times r_{2}}{2 i_{2} \times 2 r_{2}} = \frac{1}{4}$
$∴ \frac{V_{1}}{V_{2}} = \frac{4}{1}$

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