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Two concentric coils each of radius equal to $2\pi$ cm are placed at right angles to each other. 3 ampere and 4 ampere are the currents flowing in each coil respectively. The magnetic induction in weber/m² at the center of the coils will be

$(\mu_{0} =4\pi \times 10^{-7}\; Wb/A-m)$

Option 1)
$5\times 10^{-5}$

Option 2)
$7\times 10^{-5}$

Option 3)
$12\times 10^{-5}$

Option 4)
$10^{-5}$

Answers (1)

best_answer

As we learnt in

Concentric loops but their planes are perpendicular to each other -

$B = \sqrt{B_{1}^{2}+B_{2}^{2}}$

$B = \frac{\mu_{o}}{2r}\:\sqrt{i_{1}^{2}+i_{2}^{2}}$

- wherein

Magnetic induction at centre of one coil $B_{1}=\frac{\mu _{0}i_{1}}{2r}$

$Similarly\; \; \; B_{2}=\frac{\mu _{0}i_{2}}{2r}$

$\therefore \; \; \; B^{2}=B_{1}^{2}+B_{2}^{2}=\left ( \frac{\mu _{0}i_{1}}{2r} \right )^{2}+\left ( \frac{\mu _{0}i_{2}}{2r} \right )^{2}$

$=\frac{\mu _{0}\; ^{2}}{4r^{2}}(i_{1}^{2}+i_{2}^{2})$

$\therefore \; \; \: B=\frac{\mu _{0}}{2r}\sqrt{i_{1}^{2}+i_{2}^{2}}$

$=\frac{4\pi \times 10^{-7}}{2\times (2\pi \times 10^{-2})}\sqrt{(3)^{2}+(4)^{2}}$

or $B=5\times 10^{-5}\; Wb/m^{2}$

Option 1)

$5\times 10^{-5}$

Correct option

Option 2)

$7\times 10^{-5}$

Incorrect option

Option 3)

$12\times 10^{-5}$

Incorrect option

Option 4)

$10^{-5}$

Incorrect option

Posted by

Plabita

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