MHT CET · Physics · Electromagnetic Induction
Two different coils have self inductance \(8 \mathrm{mH}\) and \(2 \mathrm{mH}\). The current in both coils are increased at same constant rate. The ratio of the induced emf's in the coil is
- A \(4: 1\)
- B \(1: 4\)
- C \(1: 2\)
- D \(2: 1\)
Answer & Solution
Correct Answer
(A) \(4: 1\)
Step-by-step Solution
Detailed explanation
The induced emf, \(e=\frac{L d i}{d t}\)
Here, \(L_{1}=8 \mathrm{mH}=8 \times 10^{-3} \mathrm{H}\)
and \(L_{2}=2 \mathrm{mH}=2 \times 10^{-3} \mathrm{H}\)
So, ratio of induced emfs, \(\frac{e_{1}}{e_{2}}=\frac{L_{1} \frac{d i}{d t}}{L_{2} \frac{d i}{d t}}=\frac{L_{1}}{L_{2}}\)
(since, current is increased at same rate)
\(\Rightarrow$$\frac{\mathrm{q}}{e_{2}}=\frac{8 \times 10^{-3} \mathrm{H}}{2 \times 10^{-3} \mathrm{H}}=\frac{4}{1}\)
Here, \(L_{1}=8 \mathrm{mH}=8 \times 10^{-3} \mathrm{H}\)
and \(L_{2}=2 \mathrm{mH}=2 \times 10^{-3} \mathrm{H}\)
So, ratio of induced emfs, \(\frac{e_{1}}{e_{2}}=\frac{L_{1} \frac{d i}{d t}}{L_{2} \frac{d i}{d t}}=\frac{L_{1}}{L_{2}}\)
(since, current is increased at same rate)
\(\Rightarrow$$\frac{\mathrm{q}}{e_{2}}=\frac{8 \times 10^{-3} \mathrm{H}}{2 \times 10^{-3} \mathrm{H}}=\frac{4}{1}\)
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