Two wires of same material A and B, having lengths l and 3 2 l carry the currents I and 4 3 I respectively. Both the wires (A and B) are bent into a circles (of single turn). If B A and B B ar

Question

Two wires of same material A and B, having lengths l and 3 2 l ​ carry the currents I and 4 3 I ​ respectively. Both the wires (A and B) are bent into a circles (of single turn). If B A ​ and B B ​ are the values of magnetic field at the centres of the respective circles, then the ratio B A ​ B B ​ ​ is

TestHub · Accepted Answer

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

$B = \frac{μ _{0} I}{2 R}$

Since the wire of length $L$ is bent into a single turn circular loop, the circumference is equal to the length of the wire:

$2 π R = L \Rightarrow R = \frac{L}{2 π}$

Substituting the value of $R$ in the expression for $B$ :

$B = \frac{μ _{0} I}{2 ( \frac{L}{2 π} )} = \frac{μ _{0} π I}{L}$

For wire A:

$L_{A} = l$ and $I_{A} = I$

$B_{A} = \frac{μ _{0} π I}{l}$

For wire B:

$L_{B} = \frac{2 l}{3}$ and $I_{B} = \frac{3 I}{4}$

$B_{B} = \frac{μ _{0} π ( \frac{3 I}{4} )}{\frac{2 l}{3}} = \frac{9 μ _{0} π I}{8 l}$

Taking the ratio of the magnetic fields:

$\frac{B _{B}}{B _{A}} = \frac{\frac{9 μ _{0} π I}{8 l}}{\frac{μ _{0} π I}{l}} = \frac{9}{8}$

Answer: $\frac{9}{8}$

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