JEE Main 2014PhysicsMagnetic Effects of CurrentHardMCQ

JEE Main 2014 — Magnetic Effects of Current Question with Solution

JEE Main 2014 (06 Apr)

Question

A conductor lies along the z-axis at - 1.5 ≤ z <1.5 m  and carries a fixed current of 10.0 A in - a ^ z  direction (see figure). For a field B → = 3.0 × 1 0 - 4  e - 0.2x  a ^ y  T, find the power required to move the conductor at constant speed to x = 2.0 m, y = 0 m in 5 × 1 0 - 3  s . Assume parallel motion along the x-axis.

   

Choose an option

Show full solutionCorrect option: B
Correct answer
B2.97 W

Step-by-step explanation

Power P = W Δ t

Force on conductor
F→=I(l→×b→)

F → =-10( 3 a ^ z )×( 3× 10 −4 e −0.2x a ^ y )

⇒F→=90×10-4(e-0.2x) along with x-axis

work done on the conductor in moving along x-axis.

W=∫x=02F→.dx→

W = ∫ 0 2 Bi ℓ dx = ∫ 0 2 3 × 10 - 4 e - 0.2x × 3 × 10dx =90×10-4e-0.2x-0.202

= 450 × 10 - 4 [ 1 - e - 0.4 ] = 180 × 10 - 4 ( Using e - x = 1 - x )

P = W Δ t = 180 × 10 - 4 5 × 10 - 3 = 3.6 W.

Hence, closest option is 2.97 W.
(if we used exact value of exponential term will get exact answer).

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About this question

This is a previous-year question from JEE Main 2014, covering the Magnetic Effects of Current chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.