- Exam
- JEE Advanced
- Year
- 2016
- Paper
- Paper 2
- Subject
- Physics
- Chapter
- Electromagnetic Induction
JEE Advanced 2016 · Paper 2 · Physics · Electromagnetic Induction
A rigid wire loop of square shape having side of length L and...
A rigid wire loop of square shape having side of length L and resistance R is moving along the x-axis with a constant velocity in the plane of the paper. At t = 0, the right edge of the loop enters a region of length 3L where there is a uniform magnetic field into the plane of the paper, as shown in the figure. For sufficiently large the loop eventually crosses the region. Let x be the location of the right edge of the loop. Let v(x), I(x) and F(x) represent the velocity of the loop, current in the loop, and force on the loop, respectively, as a function of x. Counter-clockwise current is taken as positive.
Which of the following schematic plot(s) is(are) correct ? (Ignore gravity)
Options
- A
- B
- C
- D
Correct answer
C.
Step-by-step solution
When loop was entering (x < L)
\left|e\right| = BLVi = \dfrac{e}{R} = \dfrac{BLV}{R}\left(ACW\right)F = \dfrac{i}{B}\dfrac{{B}^{2}{L}^{2}V}{R}⇒ a = \dfrac{F}{m} = \dfrac{{B}^{2}{L}^{2}V}{mR}a = V\dfrac{dV}{dx}V\dfrac{dV}{dx} = \dfrac{{B}^{2}{L}^{2}V}{mR} ⇒ \mathop{∫}_{{V}_{0}}^{V}dV = - \dfrac{{B}^{2}{L}^{2}}{mR}\mathop{∫}_{0}^{x}dx⇒ V = {V}_{0} - \dfrac{{B}^{2}{L}^{2}}{mR}xI = \dfrac{BL}{R}V ⇒L \le x \le 3L<br ><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/Phy_Papr2_9_1_.jpg"/><br >\dfrac{dϕ}{dt} = 0e = 0i = 0x > 4L<br ><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/Phy_Papr2_9_2_.jpg"/><br >e = Blvi = \dfrac{BLW}{R}a = \dfrac{{B}^{2}{L}^{2}V}{mR} = V\dfrac{dV}{dx}F = \dfrac{{B}^{2}{L}^{2}V}{R} \mathop{∫}_{L}^{x}- \dfrac{{B}^{2}{L}^{2}}{mR}dx = \mathop{∫}_{{V}_{i}}^{{V}_{f}}dV⇒ \dfrac{{B}^{2}{L}^{2}}{mR}
(straight line of negative slope)
(Clockwise) (straight line of negative slope)