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Solve it, - Electromagnetic Induction and alternative Currents - NEET

A uniform magnetic field is restricted within a region of radius r. The magnetic field changes with time at a rate \frac{{d\vec B}} {{dt}}. Loop 1 of radius R > r

encloses the region r and loop 2 of radius R is outside the region of magnetic field as shown in the figure below. Then the e.m.f. generated is

 

  • Option 1)

    zero in loop 1 and zero in loop 2

  • Option 2)

    - \frac{{d\vec B}} {{dt}}\pi r^2in loop 1 and - \frac{{d\vec B}} {{dt}}\pi r^2in loop 2

  • Option 3)

    - \frac{{d\vec B}} {{dt}}\pi R^2in loop 1 and zero in loop 2

  • Option 4)

    - \frac{{d\vec B}} {{dt}}\pi r^2in loop 1 and zero in loop 2

 
Answers (1)
105 Views

As discussed in

Rate of change of magnetic Flux -

varepsilon = frac{-dphi }{dt}
 

- wherein

dphi
ightarrow phi _{2}-phi _{1}

phi _{2}-phi _{1}- change in flux

 

 

In loop 1, 

\varepsilon _{1}=\frac{-d \phi}{dt}=\frac{-d}{dt}(\vec{B}.\vec{A})=\frac{-d}{dt}(BA)=A\frac{dB}{dt}

\varepsilon _{1}= - \left( \frac{\pi r^{2}dB}{dt} \right )

 In loop 2,

\varepsilon _{2}= \frac{-d}{dt}(BA)=\frac{-d}{dt}(0 \times A)=0 


Option 1)

zero in loop 1 and zero in loop 2

This option is incorrect.

Option 2)

- \frac{{d\vec B}} {{dt}}\pi r^2in loop 1 and - \frac{{d\vec B}} {{dt}}\pi r^2in loop 2

This option is incorrect.

Option 3)

- \frac{{d\vec B}} {{dt}}\pi R^2in loop 1 and zero in loop 2

This option is incorrect.

Option 4)

- \frac{{d\vec B}} {{dt}}\pi r^2in loop 1 and zero in loop 2

This option is correct.

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