KCET · Physics · Magnetic Effects of Current
The magnetic field at the centre of a circular coil of radius \(R\) carrying current \(I\) is 64 times the magnetic field at a distance \(x\) on its axis from the centre of the coil. Then, the value of \(x\) is
- A \(\frac{R}{4} \sqrt{15}\)
- B \(R \sqrt{3}\)
- C \(\frac{R}{4}\)
- D \(R \sqrt{15}\)
Answer & Solution
Correct Answer
(D) \(R \sqrt{15}\)
Step-by-step Solution
Detailed explanation
Given, \(B_{\text {centre }}=64 \times B_{\text {axis }}\)
\(\Rightarrow \frac{\mu_0 I}{2 R}=64 \cdot \frac{\mu_0 I R^2}{2\left(R^2+x^2\right)^{\frac{3}{2}}}\)
\(\Rightarrow \left(R^2+x^2\right)^{\frac{3}{2}}=64 R^3=(4 R)^3 \Rightarrow\left(R^2+x^2\right)^{\frac{1}{2}}=4 R\)
\(\Rightarrow R^2+x^2=16 R^2 \Rightarrow 15 R^2=x^2\)
\(\Rightarrow x=\sqrt{15} R=R \sqrt{15}\)
\(\Rightarrow \frac{\mu_0 I}{2 R}=64 \cdot \frac{\mu_0 I R^2}{2\left(R^2+x^2\right)^{\frac{3}{2}}}\)
\(\Rightarrow \left(R^2+x^2\right)^{\frac{3}{2}}=64 R^3=(4 R)^3 \Rightarrow\left(R^2+x^2\right)^{\frac{1}{2}}=4 R\)
\(\Rightarrow R^2+x^2=16 R^2 \Rightarrow 15 R^2=x^2\)
\(\Rightarrow x=\sqrt{15} R=R \sqrt{15}\)
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