Magnetics 2 Question 6
6. 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. $\frac{B _L}{B _C}$ will be
(a) $\frac{1}{N}$
(b) $N$
(c) $\frac{1}{N^{2}}$
(d) $N^{2}$
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Answer:
Correct Answer: 6. (c)
Solution:
- Let consider the length of first wire is $L$, then according to question, if radius of loop formed is $R _1$, then,
For wire 1,
$$ \begin{gathered} \underset{L}{\stackrel{A}{\longrightarrow}} B \underset{R _1}{\longrightarrow} A, B \Rightarrow \\ L=2 \pi R _1 \Rightarrow R _1=\frac{L}{2 \pi} \end{gathered} $$
The magnetic field due to this loop at its centre is
$$ B _L=\frac{\mu _0 I}{2 R _1}=\frac{\mu _0 I}{2 L} \times 2 \pi $$
Now, several wire is made into a coil of
$N$ - turns.
$$ \begin{aligned} & \left(R _2=\right.\text { radius of coil } \\ & \text { having } N \text { loops) } \end{aligned} $$
Thin, $\quad L=N\left(2 \pi R _2\right) \Rightarrow R _2=\frac{L}{2 \pi N}$
The magnetic field due to this circular coil of $N$-turns is
$$ B _C=\frac{\mu _0 I}{2 R _2} N=N \cdot \frac{\mu _0 I \cdot(2 \pi N)}{2 L} $$
Using Eqs. (i) and (ii), the ratio of $\frac{B _L}{B _C}$ is :
$\Rightarrow \quad \frac{B _L}{B _C}=\frac{\frac{\mu _0 I}{2 R _1}}{N \frac{\mu _0 I}{2 R _2}}=\frac{\frac{\mu _0 I}{2 L} \cdot(2 \pi)}{\frac{\mu _0 I}{2 L} \cdot(2 \pi) N^{2}}=\frac{1}{N^{2}}$
