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PhyHardAdv 2017 • Paper 1

Straight Line Motion Along Negative Y Direction in Fields

Comprehension Passage

Answer Q.13, Q.14 and Q.15 by appropriately matching the information given in the three columns of the following table.

Consider a charged particle (electron or proton) introduced at the origin (x=0,y=0,z=0)(x = 0, y = 0, z = 0) with a given initial velocity v⃗\vec{v}. Uniform electric field E⃗\vec{E} and magnetic field B⃗\vec{B} exist everywhere. The velocity v⃗\vec{v}, electric field E⃗\vec{E}, and magnetic field B⃗\vec{B} are listed in columns 1, 2, and 3 respectively. The parameters E0E_0 and B0B_0 are positive magnitudes.

Column 1Column 2Column 3(I) Electron with v⃗=2E0B0x^(i) E⃗=E0z^(P) B⃗=−B0x^(II) Electron with v⃗=E0B0y^(ii) E⃗=−E0y^(Q) B⃗=B0x^(III) Proton with v⃗=0(iii) E⃗=−E0x^(R) B⃗=B0y^(IV) Proton with v⃗=2E0B0x^(iv) E⃗=E0x^(S) B⃗=B0z^\begin{array}{|l|l|l|} \hline \text{Column 1} & \text{Column 2} & \text{Column 3} \\ \hline \text{(I) Electron with } \vec{v} = 2\frac{E_0}{B_0}\hat{x} & \text{(i) } \vec{E} = E_0\hat{z} & \text{(P) } \vec{B} = -B_0\hat{x} \\ \hline \text{(II) Electron with } \vec{v} = \frac{E_0}{B_0}\hat{y} & \text{(ii) } \vec{E} = -E_0\hat{y} & \text{(Q) } \vec{B} = B_0\hat{x} \\ \hline \text{(III) Proton with } \vec{v} = 0 & \text{(iii) } \vec{E} = -E_0\hat{x} & \text{(R) } \vec{B} = B_0\hat{y} \\ \hline \text{(IV) Proton with } \vec{v} = 2\frac{E_0}{B_0}\hat{x} & \text{(iv) } \vec{E} = E_0\hat{x} & \text{(S) } \vec{B} = B_0\hat{z} \\ \hline \end{array}

Determine the condition where the particle moves in a straight line along the negative direction of the yy-axis (i.e., along −y^-\hat{y}).

Options

A

(II) (iii) (Q)

B

(III) (ii) (R)

Correct
C

(IV) (ii) (S)

D

(III) (ii) (P)

Topics & Concepts

Moving Charges and MagnetismMagnetic ForceElectric Field

Step-by-Step Solution

To determine the combination of initial conditions and fields that causes the charged particle to move in a straight line along the negative yy-direction (i.e., along −y^-\hat{y}), we analyze the total Lorentz force acting on the particle:

F⃗=q(E⃗+v⃗×B⃗)\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})

Let us evaluate the combination given in option (B): (III) (ii) (R):

  1. Particle parameters (Column 1, (III)):

    • Particle: Proton, so charge q=+e>0q = +e > 0.
    • Initial velocity: v⃗=0\vec{v} = 0.

  2. Electric Field (Column 2, (ii)):

    • E⃗=−E0y^\vec{E} = -E_0\hat{y}

  3. Magnetic Field (Column 3, (R)):

    • B⃗=B0y^\vec{B} = B_0\hat{y}

Step-by-Step Motion Analysis:

  • At t=0t = 0: Since the initial velocity is zero (v⃗=0\vec{v} = 0), the magnetic force acting on the particle is initially zero: F⃗m=q(v⃗×B⃗)=0\vec{F}_m = q(\vec{v} \times \vec{B}) = 0

    The electric force acting on the proton is: F⃗e=qE⃗=(+e)(−E0y^)=−eE0y^\vec{F}_e = q\vec{E} = (+e)(-E_0\hat{y}) = -eE_0\hat{y}

    The net force at t=0t=0 accelerates the particle from rest along the negative yy-axis: a⃗=−eE0my^\vec{a} = -\frac{eE_0}{m}\hat{y}

  • For t>0t > 0: As the particle accelerates along −y^-\hat{y}, its velocity at any subsequent time tt points along the negative yy-axis: v⃗(t)=vy(t)y^where vy(t)<0\vec{v}(t) = v_y(t)\hat{y} \quad \text{where } v_y(t) < 0

    The magnetic force on the particle while moving along the yy-axis is: F⃗m=q(v⃗×B⃗)=(+e)(vy(t)y^×B0y^)=0\vec{F}_m = q(\vec{v} \times \vec{B}) = (+e)\left(v_y(t)\hat{y} \times B_0\hat{y}\right) = 0

    Because the velocity vector v⃗\vec{v} is parallel to the magnetic field B⃗=B0y^\vec{B} = B_0\hat{y} at all times, the magnetic force remains zero indefinitely.

  • Conclusion: The net force on the particle remains strictly along the negative yy-axis: F⃗net=−eE0y^\vec{F}_{\text{net}} = -eE_0\hat{y}

    Therefore, the particle moves in a straight line along the negative yy-direction under constant acceleration.

Thus, the correct matching option is (B) corresponding to (III) (ii) (R).

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