Question 200: Electromagnetic Induction

Question

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 $10 \mathrm{~A} / \mathrm{s}$, the emf in coil P is 12 mV . When coil Q carries no current and current of 1.5 A flows through coil P , the magnetic flux linked with the coil Q in mWb is

Accepted Answer

Step 1: Relationship between Magnetic Flux and Mutual Inductance
The magnetic flux through coil Q because of the current in coil P is given by:

$\phi_Q = M I_P$

where $M$ is the mutual inductance, and $I_P$ is the current through coil P.
Step 2: Finding Mutual Inductance ($M$)
The emf in coil P, caused by a changing current in coil Q, is given by:

$|e_p| = M \frac{dI_Q}{dt}$

We can rearrange this formula to solve for $M$:

$M = \frac{|e_p|}{\frac{dI_Q}{dt}}$

This means you divide the emf ($e_p$) by the rate of change of current in coil Q ($\frac{dI_Q}{dt}$).
Step 3: Substitute the values
We are given:
$|e_p| = 12\,	ext{mV} = 12 	imes 10^{-3}\,	ext{V}$
$\frac{dI_Q}{dt} = 10\,	ext{A/s}$
$I_P = 1.5\,	ext{A}$
Step 4: Find $M$
Put the values into the formula:

$M = \frac{12 	imes 10^{-3}}{10} = 1.2 	imes 10^{-3}\, 	ext{H}$
Step 5: Find the Flux Linked with Coil Q
Using $\phi_Q = M I_P$:

$ \phi_Q = 1.2 	imes 10^{-3} 	imes 1.5 = 1.8 	imes 10^{-3}~	ext{Wb} $
Step 6: Convert to milliWeber (mWb)
Since $1~	ext{mWb} = 10^{-3}~	ext{Wb}$, we have:

$ \phi_Q = 1.8~	ext{mWb} $

Answer

0.9

Answer

1.2

Answer

1.5

Answer

1.8

Electromagnetic Induction

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