MHT CET · Physics · Electromagnetic Induction
The charge which will flow through a galvanometer of resistance \(200 \Omega\) connected to a \(400 \Omega\) circular coil of 1000 turns wound on a wooden stick \(20 \mathrm{~mm}\) in diameter, if a magnetic field \(\mathrm{B}=0.012 \mathrm{~T}\) parallel to the axis of the stick decreased suddenly to zero is nearly
- A \(63 \mu \mathrm{C}\)
- B \(630 \mu \mathrm{C}\)
- C \(6.3 \mu \mathrm{C}\)
- D \(0.63 \mu \mathrm{C}\)
Answer & Solution
Correct Answer
(C) \(6.3 \mu \mathrm{C}\)
Step-by-step Solution
Detailed explanation
Charge flowing through a circuit due to flux change is written as,
\(\mathrm{q}=\frac{\Delta \phi}{\mathrm{R}}=\frac{\mathrm{NA}\left(\mathrm{B}_2-\mathrm{B}_1\right)}{\mathrm{R}}=\frac{\mathrm{N} \pi \mathrm{r}^2\left(\mathrm{~B}_2-\mathrm{B}_1\right)}{\mathrm{R}}\)
On plugging the given values,
\(q=\frac{1000 \times \pi \times 10^{-4} \times(0.012-0)}{(200+400)}\)
On solving,
\(q=\frac{1000 \times \pi \times 10^{-4} \times(0.012-0)}{(200+400)}\)
\(\mathrm{q}=\frac{\Delta \phi}{\mathrm{R}}=\frac{\mathrm{NA}\left(\mathrm{B}_2-\mathrm{B}_1\right)}{\mathrm{R}}=\frac{\mathrm{N} \pi \mathrm{r}^2\left(\mathrm{~B}_2-\mathrm{B}_1\right)}{\mathrm{R}}\)
On plugging the given values,
\(q=\frac{1000 \times \pi \times 10^{-4} \times(0.012-0)}{(200+400)}\)
On solving,
\(q=\frac{1000 \times \pi \times 10^{-4} \times(0.012-0)}{(200+400)}\)
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