Electro Magnetic Induction And Alternating Currents Question 328
As a metal rod makes contact and completes the circuit. The circuit is perpendicular to the magnetic field with $ B=0.15,tesla. $ If the resistance is $ 3,\Omega $ , force needed to move the rod as indicated with a constant speed of $ 2,m/sec $ is [MP PET 1994]
Options:
A) $ 3.75\times {{10}^{-3}} $ N
B) $ 3.75\times {{10}^{-2}},N $
C) $ 3.75\times 10^{2},N $
D) $ 3.75\times {{10}^{-4}}N $
Show Answer
Answer:
Correct Answer: A
Solution:
Induced current in the circuit
$ i=\frac{Bvl}{R} $
Magnetic force acting on the wire $ F_{m}=Bil=B( \frac{Bvl}{R} )\ l $
$ \Rightarrow F_{m}=\frac{B^{2}vl^{2}}{R} $
External force needed to move the rod with constant velocity
$ (F_{m})=\frac{B^{2}vl^{2}}{R}=\frac{{{(0.15)}^{2}}\times (2)\times {{(0.5)}^{2}}}{3} $ $ =3.75\times {{10}^{-3}}N $
BETA
