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GUJCET · Physics · Electromagnetic Induction
A rod of \(1 0 ~ c m\) length is moving perpendicular to uniform magnetic field of intensity \(5 \times 10^{-4} \frac{Wb}{ m ^2}\). If the acceleration of the rod is \(5 ms^{-2}\), then the rate of increase of induced \(e m f\) is _________ .
- A \(2.5 \times 10^{-4} Vs ^{-1}\)
- B \(25 \times 10^{-4} Vs\)
- C \(20 \times 10^{-4} Vs\)
- D \(20 \times 10^{-4} Vs ^{-1}\)
Answer & Solution
Correct Answer
(A) \(2.5 \times 10^{-4} Vs ^{-1}\)
Step-by-step Solution
Detailed explanation
A
\(\varepsilon= B \cup l\)
The rate of increase of induced emf
\(\begin{array}{l}\therefore \frac{d \varepsilon}{d t}= B l \frac{d v}{d t} \\ \therefore \frac{d \varepsilon}{d t}= B l a\end{array}\)
\(\left(\because \frac{d v}{d t}=a=\right.\) acceleration \()\)
\(\begin{aligned} \therefore \quad \frac{d \varepsilon}{d t}=(5 & \left.\times 10^{-4}\right) \\ & \times\left(10 \times 10^{-2}\right) \times(5)\end{aligned}\)
\(\therefore \quad \frac{d \varepsilon}{d t}=2.5 \times 10^{-4} Vs ^{-1}\)
\(\varepsilon= B \cup l\)
The rate of increase of induced emf
\(\begin{array}{l}\therefore \frac{d \varepsilon}{d t}= B l \frac{d v}{d t} \\ \therefore \frac{d \varepsilon}{d t}= B l a\end{array}\)
\(\left(\because \frac{d v}{d t}=a=\right.\) acceleration \()\)
\(\begin{aligned} \therefore \quad \frac{d \varepsilon}{d t}=(5 & \left.\times 10^{-4}\right) \\ & \times\left(10 \times 10^{-2}\right) \times(5)\end{aligned}\)
\(\therefore \quad \frac{d \varepsilon}{d t}=2.5 \times 10^{-4} Vs ^{-1}\)
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