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JEE Advanced 2017 Paper 1 · Physics · Magnetic Effects of Current · medium

JEE Advanced 2017 Paper 1 Physics: Paragraph 2

A particle of mass mm and charge q>0q>0 moves in uniform fields. Define the time τ=πmqB0\tau=\dfrac{\pi m}{qB_0}. At t=0t=0 the particle is at the origin. Column 1 gives the fields and initial velocity, Column 2 gives the trajectory, and Column 3 gives the speed of the particle at t=τt=\tau.

Column 1: (I) B⃗=B0z^\vec B=B_0\hat z, E⃗=0\vec E=0, initial velocity v0x^v_0\hat x; (II) B⃗=B0z^\vec B=B_0\hat z, E⃗=E0y^\vec E=E_0\hat y, particle starts from rest; (III) B⃗=B0z^\vec B=B_0\hat z, E⃗=E0z^\vec E=E_0\hat z, initial velocity v0x^v_0\hat x; (IV) B⃗=0\vec B=0, E⃗=2E0y^\vec E=2E_0\hat y, initial velocity v0x^v_0\hat x.
Column 2: (i) a circle; (ii) a cycloid; (iii) a helix of increasing pitch; (iv) a parabola.
Column 3: (P) v0v_0; (Q) 2E0B0\dfrac{2E_0}{B_0}; (R) v02+(πE0B0)2\sqrt{v_0^2+\left(\dfrac{\pi E_0}{B_0}\right)^2}; (S) v02+(2πE0B0)2\sqrt{v_0^2+\left(\dfrac{2\pi E_0}{B_0}\right)^2}.

Each question asks for the one row-consistent combination, i.e. a row of Column 1 together with its correct entries from Columns 2 and 3.

Question 1

Single correct option

Which combination describes the particle whose trajectory projects on the xyxy-plane as a circle but whose kinetic energy is not conserved?
  1. A. (I) (i) (P)
  2. B. (III) (iii) (R)
  3. C. (III) (i) (R)
  4. D. (IV) (iv) (S)
Show answer and solution

Answer: (B) (III) (iii) (R)

Kinetic energy changes only if an electric field has a component along the velocity, which excludes the pure magnetic case (I). In row (III) the field E0z^E_0\hat z accelerates the particle along zz with vz=qE0t/mv_z=qE_0t/m, while B⃗∥z^\vec B\parallel\hat z keeps the transverse motion circular at constant speed v0v_0. The path is therefore a helix whose pitch grows because vzv_z increases linearly. Its projection on the xyxy-plane is a circle. At t=τt=\tau, vz=qE0mπmqB0=πE0B0v_z=\dfrac{qE_0}{m}\dfrac{\pi m}{qB_0}=\dfrac{\pi E_0}{B_0}, giving speed v02+(πE0/B0)2\sqrt{v_0^2+(\pi E_0/B_0)^2}, entry (R). The option with (i) is wrong since the real path is a helix. The answer is (III) (iii) (R).

Question 2

Single correct option

Which combination describes the particle whose speed first increases from its initial value, reaches a maximum at t=τt=\tau, and then decreases again?
  1. A. (II) (ii) (Q)
  2. B. (I) (i) (Q)
  3. C. (II) (iv) (Q)
  4. D. (II) (ii) (P)
Show answer and solution

Answer: (A) (II) (ii) (Q)

With E⃗⊥B⃗\vec E\perp\vec B and the particle starting from rest, the motion is the superposition of a uniform drift E0/B0E_0/B_0 along xx and circular motion of the same speed, so the path is a cycloid. The speed is v=2E0B0∣sin⁡ωt2∣v=\dfrac{2E_0}{B_0}\left|\sin\dfrac{\omega t}{2}\right| with ω=qB0/m\omega=qB_0/m. It rises from 00, reaches its maximum 2E0/B02E_0/B_0 when ωt=π\omega t=\pi, i.e. at t=τt=\tau, and then falls back to zero at 2τ2\tau. Thus the row is (II), the trajectory is (ii) and the speed at τ\tau is (Q). In the other rows the speed is either constant (I) or keeps increasing (III, IV). The answer is (II) (ii) (Q).

Question 3

Single correct option

Which combination describes the particle that moves in a plane and whose speed increases without limit as time goes on?
  1. A. (IV) (iv) (R)
  2. B. (II) (ii) (S)
  3. C. (IV) (iv) (S)
  4. D. (III) (iii) (S)
Show answer and solution

Answer: (C) (IV) (iv) (S)

Row (IV) has only a uniform electric field 2E0y^2E_0\hat y perpendicular to the initial velocity, so the acceleration is constant and along yy. Then x=v0tx=v_0t, y=qE0mt2y=\dfrac{qE_0}{m}t^2: the path stays in the xyxy-plane and is a parabola, entry (iv). Its speed v02+(2qE0t/m)2\sqrt{v_0^2+(2qE_0t/m)^2} grows without bound. At t=τt=\tau, vy=2qE0mπmqB0=2πE0B0v_y=\dfrac{2qE_0}{m}\dfrac{\pi m}{qB_0}=\dfrac{2\pi E_0}{B_0}, so the speed is v02+(2πE0/B0)2\sqrt{v_0^2+(2\pi E_0/B_0)^2}, entry (S). Row (II) is planar but its speed stays bounded by 2E0/B02E_0/B_0, and row (III) has speed increasing without bound but its helix is three-dimensional. The answer is (IV) (iv) (S).
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Original question written in the JEE Advanced 2017 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake