JEE Challenger
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PhyHardAdv 2017 • Paper 1

Particle Trajectory in Combined Electric and Magnetic Fields Constant Velocity

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}

Identify the set of parameters under which the particle moves along a straight line at a constant velocity.

Options

A

(III) (ii) (R)

B

(IV) (i) (S)

C

(III) (iii) (P)

D

(II) (iii) (S)

Correct

Topics & Concepts

Moving Charges and MagnetismMagnetic ForceElectric Field

Step-by-Step Solution

To find the condition under which the charged particle moves along a straight line at a constant velocity, the net Lorentz force acting on the particle must be zero:

F⃗net=q(E⃗+v⃗×B⃗)=0⃗\vec{F}_{\text{net}} = q(\vec{E} + \vec{v} \times \vec{B}) = \vec{0}

This requires the electric force and magnetic force to be equal in magnitude and opposite in direction:

E⃗+v⃗×B⃗=0⃗  ⟹  E⃗=−v⃗×B⃗\vec{E} + \vec{v} \times \vec{B} = \vec{0} \implies \vec{E} = -\vec{v} \times \vec{B}

Let me test the given options:

Option (A) (III) (ii) (R):

  • Particle: Proton (q=+eq = +e)
  • v⃗=0\vec{v} = 0
  • E⃗=−E0y^\vec{E} = -E_0\hat{y}
  • B⃗=B0y^\vec{B} = B_0\hat{y}

Since v⃗=0\vec{v} = 0, the magnetic force is zero (F⃗B=0⃗\vec{F}_B = \vec{0}), but the electric force is non-zero (F⃗E=−eE0y^≠0⃗\vec{F}_E = -eE_0\hat{y} \neq \vec{0}). Thus, the particle accelerates, so its velocity is not constant.

Option (B) (IV) (i) (S):

  • Particle: Proton (q=+eq = +e)
  • v⃗=2E0B0x^\vec{v} = 2\frac{E_0}{B_0}\hat{x}
  • E⃗=E0z^\vec{E} = E_0\hat{z}
  • B⃗=B0z^\vec{B} = B_0\hat{z}

The cross product is: v⃗×B⃗=(2E0B0x^)×(B0z^)=−2E0y^\vec{v} \times \vec{B} = \left(2\frac{E_0}{B_0}\hat{x}\right) \times (B_0\hat{z}) = -2E_0\hat{y}

The net field term is: E⃗+v⃗×B⃗=E0z^−2E0y^≠0⃗\vec{E} + \vec{v} \times \vec{B} = E_0\hat{z} - 2E_0\hat{y} \neq \vec{0}

Hence, F⃗net≠0⃗\vec{F}_{\text{net}} \neq \vec{0}.

Option (C) (III) (iii) (P):

  • Particle: Proton (q=+eq = +e)
  • v⃗=0\vec{v} = 0
  • E⃗=−E0x^\vec{E} = -E_0\hat{x}
  • B⃗=−B0x^\vec{B} = -B_0\hat{x}

Since v⃗=0\vec{v} = 0, F⃗B=0⃗\vec{F}_B = \vec{0} but F⃗E=−eE0x^≠0⃗\vec{F}_E = -eE_0\hat{x} \neq \vec{0}, so velocity is not constant.

Option (D) (II) (iii) (S):

  • Particle: Electron (q=−eq = -e)
  • v⃗=E0B0y^\vec{v} = \frac{E_0}{B_0}\hat{y}
  • E⃗=−E0x^\vec{E} = -E_0\hat{x}
  • B⃗=B0z^\vec{B} = B_0\hat{z}

Calculating the cross product: v⃗×B⃗=(E0B0y^)×(B0z^)=E0(y^×z^)=E0x^\vec{v} \times \vec{B} = \left(\frac{E_0}{B_0}\hat{y}\right) \times (B_0\hat{z}) = E_0 (\hat{y} \times \hat{z}) = E_0\hat{x}

Calculating the net Lorentz force on the electron: F⃗net=−e(E⃗+v⃗×B⃗)=−e(−E0x^+E0x^)=0⃗\vec{F}_{\text{net}} = -e \left( \vec{E} + \vec{v} \times \vec{B} \right) = -e \left( -E_0\hat{x} + E_0\hat{x} \right) = \vec{0}

Since the net force on the particle is zero, it moves in a straight line at a constant velocity v⃗=E0B0y^\vec{v} = \frac{E_0}{B_0}\hat{y}.

Correct Option: D

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