Discussion Forum

Two coils P and Q  are kept near each other. When no current flows through coil P and current increases in coil  Q at the rate 10 A/s, the emf in coil P is 15 mV. When coil Q carries no current and current of 1.8 A flows through the coil P, the magnetic flux linked with  the coil Q is 

Answer:

Explanation:

Given,  $\frac{dl_{\theta}}{dt}=10 A/s$

 $e_{p} = 15 mA$

 = 15 x 10^-3 V, l_p =1.8 A

 Induced emf in coil  P is given as,

$|e_{p}| = M.\frac{dl_{Q}}{dt}$

 Substituting the given values , we get

15 x 10^-3  =M x 10

 $\Rightarrow$   M= 15 x 10^-4 H

 magnetic fluxed with linked coil Q is givne as,

$\phi _{Q}=Ml_{p}=15  \times 10^{-4} \times 1.8$

 = $27.0 \times 10^{-4}$

 =2.7 $\times 10^{-3}$= 2.7 mWb