JEE Challenger
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PhyHardAdv 2017 • Paper 1

Helical Motion Axis along Positive Z Axis in Electromagnetic 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}

Which combination results in the particle describing a helical trajectory with its helix axis aligned with the positive zz-direction?

Options

A

(IV) (i) (S)

Correct
B

(II) (ii) (R)

C

(III) (iii) (P)

D

(IV) (ii) (R)

Topics & Concepts

Moving Charges and MagnetismMagnetic ForceElectric Field

Step-by-Step Solution

To determine which combination results in a helical trajectory with its axis aligned along the positive zz-direction, we analyze the motion of a charged particle in uniform electric (E⃗\vec{E}) and magnetic (B⃗\vec{B}) fields.

Requirements for Helical Motion along the Positive zz-Axis:

  1. Axis of the Helix: The axis of helical motion produced by a magnetic field is parallel to the direction of the magnetic field B⃗\vec{B}. Therefore, for the axis to be aligned with the zz-axis, the magnetic field must be along the zz-direction: B⃗=B0z^\vec{B} = B_0 \hat{z} From Column 3, this corresponds uniquely to entry (S).

  2. Perpendicular Velocity for Circular Motion: To form a helix, the particle must have a velocity component perpendicular to B⃗\vec{B} in the xyxy-plane. From Column 1, entry (IV) provides a proton (q=+e>0q = +e > 0) with initial velocity: v⃗=2E0B0x^\vec{v} = 2\frac{E_0}{B_0} \hat{x} Since v⃗⊥B⃗\vec{v} \perp \vec{B}, the magnetic force F⃗B=q(v⃗×B⃗)=e(2E0B0x^×B0z^)=−2eE0y^\vec{F}_B = q(\vec{v} \times \vec{B}) = e\left(2\frac{E_0}{B_0}\hat{x} \times B_0\hat{z}\right) = -2eE_0\hat{y} provides the centripetal force for uniform circular motion in the xyxy-plane.

  3. Motion along the Axis (Positive zz-Direction): From Column 2, entry (i) gives an electric field: E⃗=E0z^\vec{E} = E_0 \hat{z} The electric force acting on the positive proton is: F⃗E=qE⃗=eE0z^\vec{F}_E = q\vec{E} = eE_0 \hat{z} This force accelerates the proton continuously in the positive zz-direction with acceleration az=eE0m>0a_z = \frac{eE_0}{m} > 0.

Combining the circular motion in the xyxy-plane with linear acceleration along the positive zz-axis results in a helical trajectory whose pitch increases with time and whose axis is aligned with the positive zz-direction.

Thus, the required combination is (IV) (i) (S).

Correct Answer: A

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