Electro Magnetic Induction And Alternating Currents Question 184
Question: A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be
Options:
A) equal
B) the same
C) doubled
D) quadrupled
Show Answer
Answer:
Correct Answer: B
Solution:
-
$ P=\frac{E^{2}}{R}=\frac{\pi r^{2}}{\rho \ell }{{( \frac{d\phi }{dt} )}^{2}}=\frac{\pi r^{2}}{\rho \ell }[ \frac{d}{dt}{{(NBA)}^{2}} ] $
$ =\frac{\pi r^{2}}{\rho \ell }N^{2}A^{2}{{( \frac{dB}{dt} )}^{2}}\Rightarrow P\propto \frac{N^{2}r^{2}}{\ell } $
Case 1: $ {P1}\propto \frac{N^{2}r^{2}}{\ell } $
Case 2: $ {P2}\propto \frac{{{(4N)}^{2}}{{(r/2)}^{2}}}{4\ell } $
Note: When we decrease the radius of the wire, its length increases but volume remains the same
$\Rightarrow \frac{P1}{P2}=\frac{1}{1} $
$ \therefore $ Power remains the same.
