Question

One of the two identical conducting wires of length $L$ is bent in the form of a circular loop and the other one into a circular coil of $N$ identical turns. If the same current is passed in both, The ratio of the magnetic field at the centre of the loop $\left(B_L\right)$ to that at the centre of the coil $\left(B_C\right)$, i.e. $\frac{\left(B_L\right)}{\left(B_C\right)}$ will be

• A. $N^2$
• B. $N$
• C. $\frac{1}{N}$
• D. $\frac{1}{N^2}$

Answer 1

The correct option is D $\frac{1}{N^2}$


For loop $L=2\pi R$

$R=\text{radius of loop}\dots \left(i\right)$

$R=\frac{L}{2\pi }$

Field at centre, $B_L=\frac{{\mu }_{0}i}{2R}$

For coil $L=N\times 2\pi r$

$r=\text{radius of each turn in coil}$

$\Rightarrow R=Nr\Rightarrow r=\frac{R}{N}$ (using $\left(1\right)$)

Field at centre of coil,

$B_C=\frac{{\mu }_{0}Ni}{2r}=\frac{{\mu }_0{N}^{2}i}{2R}$

$\frac{B_L}{B_C}=\frac{1}{N^2}$

Hence, option $\left(C\right)$ is correct.