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Saturday, 31 October

JEE Main 2020 · PhysicsMultiple choiceSingle correctMediumMulti-step

JEE Main 3 September 2020, Shift 1, Physics Q14

Question 14 of 75 in this shift, Physics question 14 of 25.

An elliptical loop having resistance R, of semi major axis a, and semi minor axis b is placed in a magnetic field as shown in the figure. If the loop is rotated about the xx-axis with angular frequency ω\omega, the average power loss in the loop due to Joule heating is :

The figure, in words

Axes drawn at left: z vertical, x to the right, y slanting down-left. At right: four parallel vertical arrows labelled B (pointing up) pass through a flat ellipse drawn in the horizontal plane; the ellipse has a semi-axis marked a along the x direction (x labelled at its right end) and a semi-axis marked b along the y direction (y labelled at the lower end).
  1. (1)π2a2b2B2ω22R\frac{\pi^2 a^2 b^2 B^2 \omega^2}{2R}Official answer
  2. (2)πabBωR\frac{\pi a b B \omega}{R}
  3. (3)π2a2b2B2ω2R\frac{\pi^2 a^2 b^2 B^2 \omega^2}{R}
  4. (4)zero

Official answer

Option 1

NTA final key (Sep 2020).