Question 220: 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 of $10 \mathrm{~A} / \mathrm{S}$, the e.m.f. in coil P is 15 mV . When coil Q carries no current and current of 1.8 A flows through coil $P$, the magnetic flux linked with coil Q is

Accepted Answer

Given:
Rate of change of current in coil Q: $\dfrac{dI_Q}{dt} = 10~\mathrm{A/s}$
Induced emf in coil P: $e_P = 15~\mathrm{mV} = 15 	imes 10^{-3}~\mathrm{V}$
Current in P when Q carries no current: $I_P = 1.8~\mathrm{A}$
Flux linked with coil Q needed: $\phi_Q = ?$
Step 1: Find Mutual Inductance (M)
The mutual inductance $M$ is defined as:
$$ e_P = -M \frac{dI_Q}{dt} $$
(We can take magnitude as only magnitude is given)
$$ M = \frac{e_P}{\left|\frac{dI_Q}{dt}ight|} $$
Substitute the given values:
$$ M = \frac{15 	imes 10^{-3}}{10} = 1.5 	imes 10^{-3}~	ext{H} = 1.5~	ext{mH} $$
Step 2: Find Flux Linked with Q When P Carries Current
The total flux linked with coil Q due to current in P:
$ \phi_Q = M \cdot I_P $
Substitute the values found:
$$ \phi_Q = 1.5 	imes 10^{-3}~ 	ext{H} 	imes 1.8~	ext{A} = 2.7 	imes 10^{-3}~	ext{Wb} = 2.7~	ext{mWb} $$
Final Answer:
Option B: $2.7~	ext{mWb}$

Answer

1.8 mWb

Answer

2.7 mWb

Answer

1.5 mWb

Answer

1 mWb

Electromagnetic Induction

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