KCET · Physics · Magnetic Effects of Current
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 \(B_2=\)
- A \(B_1 / 9\)
- B \(9 B_1\)
- C \(3 B_1\)
- D \(B_1 / 3\)
Answer & Solution
Correct Answer
(A) \(B_1 / 9\)
Step-by-step Solution
Detailed explanation
Case-I
Number of turns, \(N_1=9\)
\(\therefore B_1=\frac{\mu_0 N_1 I}{2 R}=\frac{9 \mu_0 I}{2 R}\)
Case-II \(N_2=3\)
If radius of turns be \(R^{\prime}\), then
\(9 \times 2 \pi R=3 \times 2 \pi R^{\prime}\)
\(\Rightarrow R^{\prime}=3 R\)
\(\therefore B_2=\frac{\mu_0 N_2 I}{2 R^{\prime}}=\frac{\mu_0 \times 3 \times I}{2 \times 3 R}=\frac{\mu_0 I}{2 R}\)
\(\therefore \frac{B_2}{B_1}=\frac{2 R}{\frac{9 \mu_0 E}{2 R}}=\frac{1}{9} \Rightarrow B_2=\frac{B_1}{9}\)
Number of turns, \(N_1=9\)
\(\therefore B_1=\frac{\mu_0 N_1 I}{2 R}=\frac{9 \mu_0 I}{2 R}\)
Case-II \(N_2=3\)
If radius of turns be \(R^{\prime}\), then
\(9 \times 2 \pi R=3 \times 2 \pi R^{\prime}\)
\(\Rightarrow R^{\prime}=3 R\)
\(\therefore B_2=\frac{\mu_0 N_2 I}{2 R^{\prime}}=\frac{\mu_0 \times 3 \times I}{2 \times 3 R}=\frac{\mu_0 I}{2 R}\)
\(\therefore \frac{B_2}{B_1}=\frac{2 R}{\frac{9 \mu_0 E}{2 R}}=\frac{1}{9} \Rightarrow B_2=\frac{B_1}{9}\)
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