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)

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

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