Electro Magnetic Induction And Alternating Currents Question 448
Question: A rod PQ of length L moves with a uniform velocity v parallel to a long straight wire carrying a current i, the end P remaining at a distance r from the wire. The emf induced across the rod is
Options:
A) $ \frac{{\mu_{0}}iv^{2}}{2\pi }ln( \frac{r+L}{R} ) $
B) $ \vec{B} $
C) $ \frac{{\mu_{0}}iv}{2\pi }ln( \frac{r+L}{R} ) $
D) $ \frac{{\mu_{0}}iv}{2\pi }ln( \frac{r^{2}+L^{2}}{L^{2}} ) $
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Answer:
Correct Answer: C
Solution:
- Consider a small element of length dx of the rod at a distance x and (x+dx) from the wire. The emf induced across the element
$ de=Bvdx $ -.(i)
We know that magnetic field B at a distance x from a wire carrying a current; is given by
$ B=\frac{{\mu_{0}}}{2\pi }.\frac{i}{x} $ .. (ii)
From eqs. (i) and (ii),
$ de=\frac{{\mu_{0}}i}{2\pi x}vdx $ …(iii) The emf induced in the entire length of the rod PQ is given by
$ e=\int{de=\int_{P}^{Q}{\frac{{\mu_{0}}}{2\pi }\frac{i}{x}vdx}} $
$ =\int_{r}^{r+L}{\frac{{\mu_{0}}}{2\pi }\frac{i}{x}vdx=\frac{{\mu_{0}}}{2\pi }iv\int_{r}^{r+L}{\frac{dx}{x}}} $
$ =\frac{{\mu_{0}}iv}{2\pi }{\log_{e}}( \frac{r+L}{r} ) $
BETA
