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There are two long co-axial solenoids of same length l. The inner and outer coils have radii r1 and r2 and number of turns per unit length n1 and n2 respectively. The ratio of mutal inductance to the self - inductance of the inner - coil is:

 

  • Option 1)  (n1/n2
  • Option 2)  (n2/n1).(r1/r2)
  • Option 3) (n2/n1)
  • Option 4)   (n2/n1).((r2)2/(r1)2)  

Answers (1)

best_answer

 

Co efficient of self induction -

\phi \, \alpha \, I\Rightarrow N\phi \, \alpha\: I

N\phi \,=L\, I

L=\frac{N\phi }{I}\,
 

- wherein

N\phi = Number of flux linkage with coil.

 

 

Co efficient of mutual induction -

N_{2}\phi_{2} \, \alpha \, I_{1}\Rightarrow N_{2}\phi_{2} =MI_{1}
 

- wherein

N_{1}= Number of turns in primary

N_{2}= Number of turns in secondary

I_{1}= current through primary.

M= m_{o}\; \; \; \; n_{1}\: n_{2}\: \pi \: r{_{1}}^{2}

L= m_{o}\; \; \; \; n{_{1}}^{2}\: \pi \: r{_{1}}^{2}

\frac{M}{L}=\frac{n_{2}}{n_{1}}

 

 

 


Option 1)

\frac{n_{1}}{n_{2}}

Option 2)

 

\frac{n_{2}}{n_{1}}.\frac{r_{1}}{r_{2}}

 

Option 3)

 

\frac{n_{2}}{n_{1}}

Option 4)

 

\frac{n_{2}}{n_{1}}. \frac{r_{2}^{2}}{r_{1}^{2}}

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