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  • I have a doubt, kindly clarify. A circular current–carrying coil has a radius R. The distance from the centre of the coil, on the axis, where B will be 1/8 of its value at the centre of the coil is

A circular current–carrying coil has a radius R. The distance from the centre of the coil, on the axis, where B will be 1/8 of its value at the centre of the coil is

  • Option 1)

    \frac{R}{\sqrt{3}}

  • Option 2)

    \sqrt{3}R

  • Option 3)

    2\sqrt{3}R

  • Option 4)

    \frac{2\sqrt{3}}{R}

 

Answers (1)

best_answer

As we learnt in 

Magnetic Field at the axis due to circular current carrying wire -

B_{axis}=\frac{\mu_{0}}{4\pi }.\frac{2\pi Nir^{2}}{(x^{2}+r^{2})\frac{3}{2}}

- wherein

N is numbers of turn in coil

 

 B=\frac{\mu_o iR^{2}}{2(R^{2}+x^{2})^\frac{3}{2}}

B=(\frac{1}{8}).\frac{\mu_oi }{2R}   

\Rightarrow \frac{\mu_{0}iR^{2}}{2(R^{2}+x^{2})^{3/2}} = \frac{1}{8}.\frac{\mu_{0}i}{2R}

\Rightarrow 8R3 = (R2 + x2)3/2

\Rightarrow 4R2 = R2 + x2

\Rightarrow x = \sqrt{3} R


Option 1)

\frac{R}{\sqrt{3}}

incorrect

Option 2)

\sqrt{3}R

correct

Option 3)

2\sqrt{3}R

incorrect

Option 4)

\frac{2\sqrt{3}}{R}

incorrect

Posted by

prateek

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