Question

One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one ito a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop $\left(B_1\right)$ to that at the centre of the coil $\left(B_C\right),i.e.\frac{B_L}{B_C}$ will be:

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

Answer 1

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

$\text{Given:-}$ Wires of length = L

Number of identical turns = N

$\text{To find:-}$The ratio of the magnetic field at the central

of the loop to that at the centre of

the coil, i.e$\left(\frac{B_l}{B_c}\right)$

$\text{Solution:-}$

$\text{For loop L}$ = $2\pi$R $\Rightarrow R=\frac{L}{2\pi }$

$B_L=\frac{{\mu }_{0}i}{2R}=\frac{{\mu }_{0}i\pi }{L}$

$\text{For coil L}$ = N$\times 2\pi r$

$\Rightarrow r=\frac{L}{2\pi }$

$B_C=\frac{{\mu }_{0}iN}{2r}=N\times \frac{{\mu }_{0}i\pi }{L}$

So, $\frac{B_L}{B_c}=\frac{1}{N^2}$

$\text{Hence the correct option is C.}$