Question

A circular coil of wire consisting of ' $n$ ' turns each of radius $8\text{ cm}$ carries a current of $0.4\text{ A}$ . The magnitude of the magnetic field at the centre of coil is $3.14 \times 10^{-4}\text{ T}$. The value of ' $n$ ' is [Take $\mu_0 = 12.56 \times 10^{-7}\text{ SI unit}$]}

• A. $1$
• B. $10$
• C. $100$
• D. $1000$

Hint:

For magnetic field at the centre of a circular coil: \[ B=\frac{\mu_0 nI}{2R} \] Always convert radius into metres before substitution.

Answer 1

The Correct Option is C

Concept:
Magnetic field at the centre of a circular coil of \(n\) turns is: \[ B=\frac{\mu_0 n I}{2R} \] ip

Step 1: Write the given values.
\[ B=3.14\times10^{-4}\text{ T} \] \[ R=8\text{ cm}=0.08\text{ m} \] \[ I=0.4\text{ A} \] \[ \mu_0=12.56\times10^{-7} \] ip

Step 2: Solve for \(n\).
\[ n=\frac{2RB}{\mu_0 I} \] \[ n=\frac{2(0.08)(3.14\times10^{-4})}{(12.56\times10^{-7})(0.4)} \] ip

Step 3: Calculate the value.
\[ n=\frac{0.16\times 3.14\times10^{-4}}{5.024\times10^{-7}} \] \[ n=\frac{5.024\times10^{-5}}{5.024\times10^{-7}}=100 \] ip Hence, the correct answer is:
\[ \boxed{(C)\ 100} \]