Electro Magnetic Induction And Alternating Currents Question 490
Question: A current of 1.5 A flows through a solenoid of length 20.0 cm, cross-section $ 20.0cm^{2} $ and 400 turns. The current is suddenly switched off in a short time of 1.0 millisecond. Ignoring the variation in the magnetic field the ends, the average back emf induced in the solenoid is:
Options:
A) 0.3 V
B) 9.6 V
C) 30.0 V
D) 3.0 V
Show Answer
Answer:
Correct Answer: A
Solution:
-
$ e_{back}=L\frac{di}{dt} $
where $ L=\frac{{\mu_{0}}N^{2}A}{l} $
$ \therefore e_{back}=\frac{{\mu_{0}}N^{2}A}{l}[ \frac{1.5-0}{1\times 10^{3}} ] $$ =\frac{4\pi \times {{10}^{-7}}{{(400)}^{2}}\times 20\times {{10}^{-4}}}{20\times {{10}^{-2}}}\times (1.5\times 10^{3}) $ = 0.3 V
BETA
