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← JEE Advanced 2015 · Paper 1
PhysicsMultiple correct+4 / -2Question 15 of 60

JEE Advanced 2015 · Paper 1, Physics question 15

A conductor (shown in the figure) carrying constant current II is kept in the xx-yy plane in a uniform magnetic field B⃗\vec B. If FF is the magnitude of the total magnetic force acting on the conductor, then the correct statement(s) is(are)

A wire in the x-y plane carrying current I to the right: a straight segment of length L, then two bent pieces (straight sides of length R making angles π/6 and π/4 with the axis, joined by arcs of radius R) spanning a horizontal distance 2R, then another straight segment of length L. Axes: x to the right, y up.
Choose one or more options
Answer and worked solution

Official answer: A, B, C

Effective length. The force on any wire in a uniform field equals the force on a straight wire joining its ends. Here the ends are along xx, a distance L+R+R+L=2(L+R)L+R+R+L=2(L+R) apart:
F⃗=I (2(L+R)x^)×B⃗.\vec F=I\,(2(L+R)\hat x)\times\vec B.

  • (A) B⃗∥z^\vec B\parallel\hat z: F=2I(L+R)B∝(L+R)F=2I(L+R)B\propto(L+R). True.
  • (B) B⃗∥x^\vec B\parallel\hat x: F=0F=0. True.
  • (C) B⃗∥y^\vec B\parallel\hat y: F=2I(L+R)B∝(L+R)F=2I(L+R)B\propto(L+R). True.
  • (D) False.

(A), (B), (C)

Topics:Force on a current-carrying conductor
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