Electro Magnetic Induction And Alternating Currents Question 328

Question: As a metal rod makes contact and complete 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 $ 2m/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 $

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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 $



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