Electro Magnetic Induction And Alternating Currents Question 344
Question: How much length of a very thin wire is required to obtain a solenoid of length $ l_{0} $ and inductance L
Options:
A) $ \sqrt{\frac{2\pi Ll_{0}}{{\mu_{0}}}} $
B) $ \sqrt{\frac{4\pi Ll_{0}}{\mu _{0}^{2}}} $
C) $ \sqrt{\frac{4\pi Ll_{0}}{{\mu_{0}}}} $
D) $ \sqrt{\frac{8\pi Ll_{0}}{{\mu_{0}}}} $
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Answer:
Correct Answer: C
Solution:
Suppose solenoid has N turns, each of radius r and length of wire is l. It’s self inductance $ L=\frac{{\mu_{0}}N^{2}A}{l_{0}}=\frac{{\mu_{0}}N^{2}\pi r^{2}}{l_{0}} $ …. (i) Also length of the wire $ l=N\times 2\pi r $
$ \Rightarrow N^{2}r^{2}=\frac{l^{2}}{4{{\pi }^{2}}} $ …. (ii) From equation (i) and (ii) $ l=\sqrt{\frac{4\pi Ll_{o}}{{\mu_{o}}}} $
BETA
