Two circular coils $P$ and $Q$ are made of two identical wires of the same length. The number of turns in the $P$ and $Q$ are 4 and 2, respectively. The magnetic inductions at the centre of $P$ and $Q$ are $B_P$ and $B_Q$, respectively. The ratio $\frac{B_P}{B_Q}$ is

• A

  0.25
• Correct answer
  B

  4
• C

  0.5
• D

  2

Solution

Let the two coil $P$ and $Q$ having same length and number of turns $N_P=4$ and $N_Q=2$ Then, $l=N_P 2 \pi R_P=N_Q 2 \pi R_Q$ where, $R_P=$ radius of coil $P$. and $\quad R_Q=$ radius of $\operatorname{coil} Q$.
If same current is passing through $P$ and $Q$. $\frac{B_P}{B_a}=\frac{\frac{N_P \mu_0 i}{2 R_P}}{\frac{N_Q \mu_0 i}{2 R_Q}}=\frac{R_Q N_P}{R_P N_Q}=2 \times 2 \Rightarrow \frac{B_P}{B_Q}=4$

Hence, the correct answer is option: 'B'.