Consider the motion of a positive point charge in a region where there are simultaneous uniform elec

Q: Why doesn't the magnetic field change the pitch?

The magnetic field only exerts a force perpendicular to the velocity, changing its direction but not its magnitude. The pitch depends on the velocity parallel to the magnetic field, which is only affected by the electric field.

Q: What exactly is the pitch in helical motion?

The pitch is the linear distance traveled along the axis of the helix during one complete revolution. If the parallel velocity is constant, the pitch is constant. If it's accelerating, the pitch increases.

Q: Why is the magnetic force zero when $\theta = 90^\circ$?

When $\theta = 90^\circ$, the velocity is entirely along the $y$-axis, which is parallel to the magnetic field. The magnetic force is given by the cross product $\mathbf{v} \times \mathbf{B}$, and the cross product of parallel vectors is always zero.

Q: Can a magnetic field ever do work on a charge?

No, the magnetic force is always perpendicular to the instantaneous velocity of the charge. Since work is the dot product of force and displacement, and they are perpendicular, the work done by a magnetic field is always zero.

Q: What would happen if the electric field was zero?

Without the electric field, there would be no acceleration along the $y$-axis. The particle would move in a helix with a constant pitch (if $\theta \neq 0^\circ$) or a pure circle in the $x$-$z$ plane (if $\theta = 0^\circ$).

Question

Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields $\mathbf{E} = E_0 \hat{\mathbf{j}}$ and $\mathbf{B} = B_0 \hat{\mathbf{j}}$. At time $t = 0$, this charge has velocity $\mathbf{v}$ in the $x$-$y$ plane, making an angle $	heta$ with the $x$-axis. Which of the following option(s) is(are) correct for time $t > 0$?

Options:

(A) If $	heta = 0^\circ$, the charge moves in a circular path in the $x$-$z$ plane.

(B) If $	heta = 0^\circ$, the charge undergoes helical motion with constant pitch along the $y$-axis.

(C) If $	heta = 10^\circ$, the charge undergoes helical motion with its pitch increasing with time, along the $y$-axis.

(D) If $	heta = 90^\circ$, the charge undergoes linear but accelerated motion along the $y$-axis.

Sigma Percentile · Accepted Answer

Imagine a charged particle entering a region of space where both electric and magnetic fields are present, perfectly aligned along the same axis. It's a classic dance of forces, a beautiful interplay between linear acceleration and circular motion. In this problem, we are tasked with predicting the exact trajectory of a positive point charge under the simultaneous influence of a uniform electric field $\mathbf{E} = E_0 \hat{\mathbf{j}}$ and a uniform magnetic field $\mathbf{B} = B_0 \hat{\mathbf{j}}$.

### The Master Equation: Lorentz Force

To understand the motion, we must look at the master equation governing charged particles in electromagnetic fields: the Lorentz force. The total force $\mathbf{F}$ is the vector sum of the electric force and the magnetic force:

$$\mathbf{F} = q\mathbf{E} + q(\mathbf{v} 	imes \mathbf{B})$$

This equation is incredibly powerful because it allows us to decouple the motion. The electric force, $\mathbf{F}_e = qE_0 \hat{\mathbf{j}}$, acts purely along the $y$-axis. It doesn't care about the particle's velocity; it simply provides a constant acceleration in the $y$-direction.

On the other hand, the magnetic force, $\mathbf{F}_m = q(\mathbf{v} 	imes \mathbf{B})$, is a bit more selective. Because the magnetic field is along the $y$-axis, the cross product dictates that the magnetic force will only act on the component of velocity that is *perpendicular* to the $y$-axis (i.e., the $x$-component). This force will always be perpendicular to both the velocity and the magnetic field, forcing the particle into a circular path in the $x$-$z$ plane.

### Analyzing the Cases

Let's break down the options by analyzing the initial angle $	heta$ the velocity vector makes with the $x$-axis.

**Case 1: $	heta = 0^\circ$**
If $	heta = 0^\circ$, the initial velocity is entirely along the $x$-axis ($\mathbf{v} = v \hat{\mathbf{i}}$). The magnetic field grabs this perpendicular velocity and starts bending it into a circle in the $x$-$z$ plane. If there were no electric field, the particle would just trace a perfect circle forever. However, the electric field is relentlessly pushing the particle along the $y$-axis, accelerating it. The combination of circular motion in the $x$-$z$ plane and accelerated linear motion along the $y$-axis results in a **helical path with an increasing pitch**. Therefore, options (a) and (b) are incorrect.

**Case 2: $	heta = 10^\circ$**
Now, the velocity has both an $x$-component ($v \cos 10^\circ$) and a $y$-component ($v \sin 10^\circ$). The physics remains fundamentally the same! The $x$-component is perpendicular to the magnetic field, so it initiates the circular motion in the $x$-$z$ plane. The $y$-component is parallel to the magnetic field, so it ignores the magnetic field completely. But the electric field catches it, accelerating the particle further along the $y$-axis. Once again, we get a **helical path with an increasing pitch**. This makes option (c) absolutely correct.

**Case 3: $	heta = 90^\circ$**
Finally, consider $	heta = 90^\circ$. The velocity is now entirely along the $y$-axis ($\mathbf{v} = v \hat{\mathbf{j}}$). This means the velocity is perfectly parallel to the magnetic field. What happens to the magnetic force? The cross product of parallel vectors is zero!

$$\mathbf{F}_m = q(v \hat{\mathbf{j}} 	imes B_0 \hat{\mathbf{j}}) = 0$$

The magnetic field becomes completely invisible to the particle. The only force acting on it is the electric force along the $y$-axis. Consequently, the particle will undergo **linear, accelerated motion** straight along the $y$-axis. This confirms that option (d) is also correct.

### The Takeaway

When dealing with combined electric and magnetic fields, the secret is to always resolve the velocity into components parallel and perpendicular to the magnetic field. The perpendicular component dances with the magnetic field to create circles, while the parallel component rides the electric field to create linear acceleration. Combine them, and you have the full picture!

Select Answer:

Visualized Solution

$Initial Setup$

$Lorentz Force$

$Case 1: θ = 0^{\circ}$

$Case 2: θ = 1 0^{\circ}$

$Case 3: θ = 9 0^{\circ}$

$Final Answer$

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Master Equation

Analyzing the Cases

The Takeaway

Similar Questions

Comprehension Passage

In which case would the particle move in a straight line along the negative direction of Y-axis (i.e. move along $- \overset{y}{^}$ )?

In which case will the particle move in a straight line with constant velocity?

In which case will the particle describe a helical path with axis along the positive z-direction?

In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a

For a positively charged particle moving in a $x - y$ plane initially along the $x$ -axis, there is a sudden change in its path due to the presence of electric and/or magnetic fields beyond $P$ . The curved path is shown in the $x - y$ plane and is found to be non-circular.

A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a

An electron moving with a speed $u$ along the positive $x$ -axis at $y = 0$ enters a region of uniform magnetic field $B = - B_{0} \hat{k}$ which exists to the right of $y$ -axis. The electron exits from the region after sometime with the speed $v$ at coordinate $y$ , then

A charged particle with charge $q$ enters a region of constant, uniform and mutually orthogonal fields $E$ and $B$ with a velocity $v$ perpendicular to both $E$ and $B$ and comes out without any change in magnitude or direction of $v$ . Then,