Question

Two wires $A$ and $B$ have the same length equal to $44cm$ and carry a current of $10A$ each. Wire $A$ is bent into a circle and wire $B$ into a square. (a) Which wire produces a greater magnetic field at the centre? (b) Obtain the magnitudes of the field at the centres of the two wires.

Answer 1

Given that,

Length = $44cm$

Current = $10A$

Now, for circle

the length is bent in form of circle then the circumference is

$2\pi r=44$

$r=\frac{44}{2\pi }$

Now, the magnetic field is

$B_A=\frac{{\mu }_{0}I}{2\pi r}$

$B_A=\frac{4\pi \times {10}^{-7}\times 10\times 2\pi }{2\times 44\times {10}^{-2}}$

$B_A=0.89\times {10}^{-4}T$

Now, for square

Length $44cm$ is form in square

So, each side has length $11cm$

Now, the magnetic field due to first wire at the center

$B=\frac{{\mu }_{0}I}{4\pi a}(\cos {45}^0+\cos {45}^0)$

Now, due to the all four wires

$B_B=\frac{4{\mu }_{0}I}{4\pi a}(\cos {45}^0+\cos {45}^0)$

$B_B=\frac{4\times 4\pi \times {10}^{-7}\times 10\times 2\times 2}{4\pi \times 11\times \sqrt{2}\times {10}^{-2}}$

$B_B=1.02\times {10}^{-4}T$

Hence, the square loop will have more magnetic field at the center