A particle of mass m and charge q is in an electric and magnetic field is given by E = 2i + 3j, B =

Q: Why is the work done by the magnetic field zero?

The magnetic force is given by $q(\mathbf{v} \times \mathbf{B})$. By the definition of the cross product, this force is always perpendicular to the velocity $\mathbf{v}$. Since work is the dot product of force and displacement ($d\mathbf{r} = \mathbf{v}dt$), the dot product of perpendicular vectors is zero.

Q: Does the shape of the path matter for the electric field work?

No! A uniform electric field is a conservative field. This means the work done depends only on the initial and final positions, not on the specific path taken between them.

Q: What if the magnetic field was changing with time?

If the magnetic field were time-varying, it would induce an additional electric field (according to Faraday's Law), which could do work. However, in this problem, the magnetic field is static.

Q: Why did we ignore the z-component in our displacement?

The particle moves from $(0,0,0)$ to $(1,1,0)$. Since there is no change in the z-coordinate, $dz = 0$, and the displacement vector lies entirely in the xy-plane.

Q: The book solution says $\mathbf{v} \times \mathbf{B} = 0$, is that true?

No, that is a typo in the reference material. The cross product $\mathbf{v} \times \mathbf{B}$ is not zero here. However, the *work done* by that force is zero because the force is perpendicular to the displacement.

Question

A particle of mass $m$ and charge $q$ is in an electric and magnetic field is given by $\mathbf{E} = 2\hat{i} + 3\hat{j}$, $\mathbf{B} = 4\hat{j} + 6\hat{k}$. The charged particle is shifted from the origin to the point $P(x=1; y=1)$ along a straight path. The magnitude of the total work done is

Options:

(A) $(0.35)q$

(B) $(0.15)q$

(C) $(2.5)q$

(D) $5q$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are tracking a charged particle as it travels through a region filled with both an electric and a magnetic field. The particle starts its journey at the origin $O(0,0)$ and moves along a straight path to the point $P(1,1)$.

We are given the electric field vector $\mathbf{E} = 2\hat{i} + 3\hat{j}$ and the magnetic field vector $\mathbf{B} = 4\hat{j} + 6\hat{k}$. Our mission is to find the total work done by the electromagnetic field on this particle during its journey.

### The Master Equation: Lorentz Force

To find the work done, we first need to know the total force acting on the particle. In an electromagnetic field, this is given by the Lorentz force equation:

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

This force has two distinct components: the electric force $\mathbf{F}_e = q\mathbf{E}$ and the magnetic force $\mathbf{F}_m = q(\mathbf{v} 	imes \mathbf{B})$.

### The Magnetic Force "Catch"

Here is a beautiful and crucial concept in physics: **the magnetic force never does any work on a moving charge.**

Why? Because the magnetic force $q(\mathbf{v} 	imes \mathbf{B})$ is always perfectly perpendicular to the velocity vector $\mathbf{v}$. Work is defined as the dot product of force and displacement ($dW = \mathbf{F} \cdot d\mathbf{r}$). Since displacement $d\mathbf{r} = \mathbf{v} dt$, the dot product of perpendicular vectors is always zero.

$$W_m = \int q(\mathbf{v} 	imes \mathbf{B}) \cdot (\mathbf{v} dt) = 0$$

*(Note: Some reference books mistakenly claim that $\mathbf{v} 	imes \mathbf{B} = 0$ for this problem. That is mathematically incorrect. The cross product is non-zero, but its dot product with the displacement is what makes the work zero!)*

### Calculating the Electric Work

Since the magnetic field is lazy and does no work, the entire burden falls on the electric field. The total work done is simply the line integral of the electric force:

$$W = \int \mathbf{F}_e \cdot d\mathbf{r} = \int q\mathbf{E} \cdot d\mathbf{r}$$

Let's substitute our given electric field $\mathbf{E} = 2\hat{i} + 3\hat{j}$ and the infinitesimal displacement vector $d\mathbf{r} = dx\hat{i} + dy\hat{j}$:

$$dW = q(2\hat{i} + 3\hat{j}) \cdot (dx\hat{i} + dy\hat{j})$$

Taking the dot product (multiplying the $\hat{i}$ components together and the $\hat{j}$ components together), we get:

$$dW = q(2dx + 3dy)$$

### Final Calculation

Now, we just need to integrate this expression from our starting point $(0,0)$ to our destination $(1,1)$. The $x$-coordinate goes from $0$ to $1$, and the $y$-coordinate goes from $0$ to $1$.

$$W = q \int_0^1 2dx + q \int_0^1 3dy$$

Evaluating these simple integrals:

$$W = q[2x]_0^1 + q[3y]_0^1$$

$$W = 2q(1 - 0) + 3q(1 - 0)$$

$$W = 2q + 3q = \mathbf{5q}$$

And there we have it! The total work done on the particle is elegantly $5q$.

Select Answer:

Visualized Solution

$Analyzing the Problem$

$Lorentz Force$

$Work Done by Magnetic Field$

$Work Done by Electric Field$

$Substituting the Vectors$

$Dot Product$

$Integration Limits$

$Final Calculation$

The Sigma Insight: Motion of a Charge in Magnetic Fields

Analyzing the Setup

The Master Equation

The Magnetic Force "Catch"

Calculating the Electric Work

Final Calculation

Similar Questions

A charge $Q$ is moving $d l$ distance in the magnetic field $B$ . Find the value of work done by $B$ .

A particle of mass $M$ and charge $Q$ moving with velocity $v$ describes a circular path of radius $R$ when subjected to a uniform transverse magnetic field of induction $B$ . The work done by the field when the particle completes one full circle is

An electric charge $+ q$ moves with velocity $v = 3 \hat{i} + 4 \hat{j} + \hat{k}$ , in an electromagnetic field given by $E = 3 \hat{i} + \hat{j} + 2 \hat{k}$ , $B = \hat{i} + \hat{j} - 3 \hat{k}$ . The $y$ component of the force experienced by $+ q$ is

A particle of mass $m$ and charge $q$ has an initial velocity $v = v_{0} \hat{j}$ . If an electric field $E = E_{0} \hat{i}$ and magnetic field $B = B_{0} \hat{i}$ act on the particle, its speed will double after a time

The region between $y = 0$ and $y = d$ contains a magnetic field $B = B \hat{k}$ . A particle of mass $m$ and charge $q$ enters the region with a velocity $v = v \hat{i}$ . If $d = \frac{m v}{2 q B}$ , then the acceleration of the charged particle at the point of its emergence at the other side is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Problem

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lorentz Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Work Done by Magnetic Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Work Done by Electric Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting the Vectors

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dot Product

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integration Limits

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation