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JIPMER 2016
Exam
JIPMER
Year
2016
Subject
Physics
Chapter
Magnetic Effects of Current

JIPMER 2016 · Physics · Magnetic Effects of Current

A circular current carrying coil has a radius R . The distance from...

A circular current carrying coil has a radius RR. The distance from the centre of the coil, on the axis, where BB will be 18\frac{1}{8} of its value at the centre of the coil is

Options

  1. AR3\frac{R}{\sqrt{3}}
  2. B3R\sqrt{3} R
  3. C23R2 \sqrt{3} R
  4. D2R3\frac{2 R}{\sqrt{3}}

Correct answer

B. 3R\sqrt{3} R

Step-by-step solution

B=μ0NiR22(R2+x2)3/2B=\frac{\mu_0 N i R^2}{2\left(R^2+x^2\right)^{3 / 2}}=μ0Ni2R⋅1[1+x2R2]3/2=BC⋅1(1+x2R2)3/2[ Let BC=μ0Ni2R]\begin{aligned} & =\frac{\mu_0 N i}{2 R} \cdot \frac{1}{\left[1+\frac{x^2}{R^2}\right]^{3 / 2}} \\ & =B_C \cdot \frac{1}{\left(1+\frac{x^2}{R^2}\right)^{3 / 2}} \quad\left[\text{ Let } B_C=\frac{\mu_0 N i}{2 R}\right]\end{aligned}⇒BC8=BC(1+x2R2)3/2[∵B=BC8]\Rightarrow \quad \frac{B_C}{8}=\frac{B_C}{\left(1+\frac{x^2}{R^2}\right)^{3 / 2}} \quad\left[\because B=\frac{B_C}{8}\right]⇒(1+x2R2)3/2=8⇒1+x2R2=4⇒x2R2=3∴x=R3\begin{array}{rlrl}\Rightarrow & \left(1+\frac{x^2}{R^2}\right)^{3 / 2} & =8 \\ \Rightarrow & 1+\frac{x^2}{R^2} & =4 \\ \Rightarrow & \frac{x^2}{R^2} & =3 \\ & \therefore & x & =R \sqrt{3}\end{array}$$
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