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  • Two coaxial loops of different radii such as <img alt="5.2 \mathrm{~cm}" s

Two coaxial loops L_1\; \&\; L_2 of different radii such as 5.2 \mathrm{~cm} and 6.4 \mathrm{~cm} are placed as shown in the figure. What should be the magnitude and direction of the current in the loop  \mathrm{L}_2 so that the net magnetic field at the point O be zeno?

Option: 1

0.54A


Option: 2

0.66A


Option: 3

0.68A


Option: 4

0.35A


Answers (1)

best_answer

The resultant magnetic field at the point ' $O$ ' will be zeno only if the direction of current in loop L_2 is opposite to that on loop L_1. Then magnetic field at O due to current I_1 in loop L_1= magnetic field at 'O' dur to current I_2 in loop L_2.

\frac{\mu_0}{4 \pi} \frac{2 \pi I_1 a_1^2}{\left(a_1^2+x_1^2\right)^{3 / 2}}=\frac{\mu_0}{4 \pi} \frac{2 \pi I_2 a_2^2}{\left(a_2^2+x_2^2\right)^{3 / 2}}

\Rightarrow \frac{I_1 a_1^2}{\left(a_1^2+x_1^2\right)^{3 / 2}}=\frac{I_2 a_2^2}{\left(a_2^2+x_2^2\right)^{3 / 2}}

\Rightarrow \frac{1 \times\left(5.2 \times 10^{-2}\right)^2}{\left[\left(5.2 \times 10^{-2}\right)^2+\left(6.4 \times 10^{-2}\right)^2\right] 3 / 2}

=\frac{I_2\left[6.4 \times 10^{-2}\right]^2}{\left[\left(6.4 \times 10^{-2}\right)^2+\left(5.2 \times 10^{-2}\right)^2\right]^{3 / 2}}

\begin{aligned} \Rightarrow \quad I_2 & =\frac{\left[5.2 \times 10^{-2}\right]^2}{\left[6.4 \times 10^{-2}\right]^2} \\ & =\frac{27.04 \times 10^{-4}}{40.96 \times 10^{-4}} \\ & =0.66 \mathrm{~A} \end{aligned}

Posted by

seema garhwal

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