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  • Please,please help me - Electromagnetic Induction and Alternating currents - JEE Main-2

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross­sectional area A = 10 cm2 and length = 20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is

(\mu _{0}=4\pi \times 10^{-7}\, TmA^{-1})

  • Option 1)

    2.4\pi \times 10^{-4}H

  • Option 2)

    2.4\pi \times 10^{-5}H

  • Option 3)

    4.8\pi \times 10^{-4}H

  • Option 4)

    4.8\pi \times 10^{-5}H

 

Answers (1)

best_answer

As we learnt in 

Mutual Induction -

Whenever the current passing throug the coil or circuit changes, the magnetic flux linked with neighbouring coil or circuit will also change.

- wherein

M=\mu _{0}n_{1}n_{2}\pi r_{1}\, ^{2}l.

From\; \phi _{2}=\pi r_{1}\, ^{2}(\mu _{0}ni)n_{2}l

A=\pi r_{1}\, ^{2}=10cm^{2},l=20cm,n_{1}=300,n_{2}=400

M=\frac{\mu _{0}n_{1}n_{2}A}{l}=\frac{4\pi \times 10^{-7}\times 300\times 400\times 10\times 10^{-4}}{0.20}

=2.4\pi \times 10^{-4}H

 


Option 1)

2.4\pi \times 10^{-4}H

This option is correct

Option 2)

2.4\pi \times 10^{-5}H

This option is incorrect

Option 3)

4.8\pi \times 10^{-4}H

This option is incorrect

Option 4)

4.8\pi \times 10^{-5}H

This option is incorrect

Posted by

divya.saini

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