Q138 / 180 · KCET 2024 · Physics

A coil having 9 turns carrying a current produces magnetic field $B_1$ at the centre. Now the coil is rewounded into 3 turns carrying same current. Then, the magnetic field at the centre $B_2=$

  • Correct answer
    A
    $B_1 / 9$
  • B
    $9 B_1$
  • C
    $3 B_1$
  • D
    $B_1 / 3$

Solution

Case-I Number of turns, $N_1=9$ $\therefore \quad B_1=\frac{\mu_0 N_1 I}{2 R}=\frac{9 \mu_0 I}{2 R}$ Case-II $N_2=3$ If radius of turns be $R^{\prime}$, then $9 \times 2 \pi R=3 \times 2 \pi R^{\prime}$ $\Rightarrow \quad R^{\prime}=3 R$ $\therefore \quad B_2=\frac{\mu_0 N_2 I}{2 R^{\prime}}=\frac{\mu_0 \times 3 \times I}{2 \times 3 R}=\frac{\mu_0 I}{2 R}$ $\therefore \quad \frac{B_2}{B_1}=\frac{2 R}{\frac{9 \mu_0 E}{2 R}}=\frac{1}{9} \Rightarrow B_2=\frac{B_1}{9}$

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

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