Electro Magnetic Induction And Alternating Currents Question 473
Question: A copper rod of length 0.19 m is moving parallel to a long wire with a uniform velocity of 10 m/s. The long wire carries 5 ampere current and is perpendicular to the rod. The ends of the rod are at distances 0.01 m and 0.2 m from the wire. The emf induced in the rod will be-
Options:
A) $ 10,\mu V $
B) $ 20,\mu V $
C) $ 30,\mu V $
D) $ 40,\mu V $
Show Answer
Answer:
Correct Answer: C
Solution:
-
EMF induced in an element of length dx at a distance x from wire = $ Bvdx $
$ \therefore $ Total EMF induced in the rod $ E=\int\limits_{0.01}^{0.2}{Bv}dx=\int\limits_{0.01}^{0.2}{\frac{{\mu_{0}}iv}{2\pi x}} $$ dx=\frac{{\mu_{0}}iv}{2\pi } $
$ \int\limits_{0.01}^{0.2}{\frac{1}{x}},dx $
$ E=\frac{\mu_0iv}{2\pi }[ \log_ex ]_{0.01}^{0.2}$=
$\frac{\mu_{0}iv}{2\pi }[\log_10(0.2)-{\log_{10}} $ $ (0.01)]\times 2.303 $
$ E=\frac{4\pi \times {{10}^{-7}}\times 5\times 10}{2\pi }[1.301]\times 2.303 $
$ =2.99\times {{10}^{-5}},V\approx 30\mu V $
BETA
