A wire of resistance 4 Ω is used to make a coil of radius 7 cm. The wire has a diameter of 1.4 mm and the resistivity of its material is 2 x 10−7 Ω m. The number of turns in the coil will be

The correct option is 70.

Let us understand how to find the number of turns in the coil step-by-step.

Step 1: Identify the given values from the problem statement
Resistance of the wire, $R=4 \Omega$
Radius of the coil, $r=7 \text{cm}=0.07 \text{m}$
Diameter of the wire, $d=1.4 \text{mm}=1.4\times {10}^{-3} \text{m}$
Resistivity of the material of the wire, $\rho =2\times {10}^{-7} \Omega \text{m}$

Step 2: Calculate the radius and cross-sectional area of the wire
The radius of the wire $r_w$ is half of its diameter:
$r_w=\frac{d}{2}=\frac{1.4\times {10}^{-3}}{2}=0.7\times {10}^{-3} \text{m}=7\times {10}^{-4} \text{m}$

The cross-sectional area $A$ of the wire is given by the formula:
$A=\pi r_w^2$

Substituting the value of $r_w$ and using $\pi \approx \frac{22}{7}$:
$A=\frac{22}{7}\times {(7\times {10}^{-4})}^2=\frac{22}{7}\times 49\times {10}^{-8}=154\times {10}^{-8} {\text{m}}^2$

Step 3: Relate resistance to the length of the wire
The resistance $R$ of a wire is related to its resistivity $\rho$, length $L$, and cross-sectional area $A$ by the formula:
$R=\frac{\rho L}{A}$

Rearranging this formula to solve for the total length $L$ of the wire:
$L=\frac{RA}{\rho }$

Substituting the values of $R$, $A$, and $\rho$:
$L=\frac{4\times 154\times {10}^{-8}}{2\times {10}^{-7}}$

Simplifying the expression:
$L=2\times 154\times {10}^{-1}=308\times 0.1=30.8 \text{m}$

Step 4: Find the number of turns in the coil
When the wire is wound into a circular coil of radius $r$ with $N$ turns, the total length $L$ of the wire is equal to $N$ times the circumference of a single turn:
$L=N\times \left(2\pi r\right)$

Therefore, the number of turns $N$ is:
$N=\frac{L}{2\pi r}$

Substituting the values of $L$, $\pi \approx \frac{22}{7}$, and $r=0.07 \text{m}$:
$N=\frac{30.8}{2\times \frac{22}{7}\times 0.07}$

Simplifying the denominator:
$2\times \frac{22}{7}\times 0.07=44\times 0.01=0.44$

Now, calculate $N$:
$N=\frac{30.8}{0.44}=70$

Thus, the number of turns in the coil is 70.