A charged particle carrying charge 1 uC is moving with velocity (2i + 3j + 4k) m/s. If an external m

Q: Why is there no electric force acting on the particle?

The problem only specifies an external magnetic field. Unless an electric field is explicitly mentioned, we assume it is zero, so the Lorentz force reduces to just the magnetic component.

Q: How do I handle the $10^{-3}$ in the magnetic field vector?

You can factor it out of the vector entirely. When setting up the determinant, it's a scalar multiplier that can be pulled out to the front, combining with the $10^{-6}$ from the charge.

Q: I keep messing up the signs when expanding the determinant. Any tips?

Always remember the alternating signs for the cofactors: $+ \hat{i}$, $- \hat{j}$, $+ \hat{k}$. For the $\hat{j}$ component, calculate the $2 \times 2$ determinant as usual, but don't forget to multiply the entire result by $-1$.

Q: What does the $\mathbf{F} \times 10^{-9}$ mean in the question?

It's a way to format the answer. The actual force is a very small number. By writing it as $\mathbf{F} \times 10^{-9}$, the question is asking you to find the vector $\mathbf{F}$ which represents the significant digits of the force.

Q: Is the force vector always perpendicular to the velocity?

Yes! The cross product $\mathbf{v} \times \mathbf{B}$ mathematically guarantees that the resulting vector is perpendicular to both $\mathbf{v}$ and $\mathbf{B}$. This is why magnetic forces can change a particle's direction but never its speed.

Question

A charged particle carrying charge $1 \mu	ext{C}$ is moving with velocity $(2\hat{i} + 3\hat{j} + 4\hat{k}) 	ext{ ms}^{-1}$. If an external magnetic field of $(5\hat{i} + 3\hat{j} - 6\hat{k}) 	imes 10^{-3} 	ext{ T}$ exists in the region, where the particle is moving, then the force on the particle is $\mathbf{F} 	imes 10^{-9} 	ext{ N}$. The vector $\mathbf{F}$ is

Options:

(A) $-0.30\hat{i} + 0.32\hat{j} - 0.09\hat{k}$

(B) $-30\hat{i} + 32\hat{j} - 9\hat{k}$

(C) $-300\hat{i} + 320\hat{j} - 90\hat{k}$

(D) $-3.0\hat{i} + 3.2\hat{j} - 0.9\hat{k}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a charged particle zipping through space. We are given its velocity vector $\mathbf{v} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and the external magnetic field $\mathbf{B} = (5\hat{i} + 3\hat{j} - 6\hat{k}) 	imes 10^{-3} 	ext{ T}$ present in that region. The charge of the particle is $q = 1 \mu	ext{C} = 10^{-6} 	ext{ C}$. Our goal is to find the magnetic force acting on this particle.

### The Master Equation

When a charge moves in a magnetic field, it experiences a magnetic Lorentz force. The fundamental equation governing this interaction is:

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

This elegant cross product tells us a profound physical truth: the magnetic force is always perfectly perpendicular to both the particle's velocity and the magnetic field. Because it acts at a right angle to the motion, a magnetic field can steer a particle but can never speed it up or slow it down!

### Setting up the Calculation

Let's substitute our known values into the formula. To compute the cross product of two vectors, we set up a $3 	imes 3$ determinant:

$$\mathbf{F} = 10^{-6} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 5 	imes 10^{-3} & 3 	imes 10^{-3} & -6 	imes 10^{-3} \end{vmatrix}$$

To make the arithmetic cleaner, notice that $10^{-3}$ is a common factor in the third row. We can pull it out of the determinant and combine it with the $10^{-6}$ outside:

$$\mathbf{F} = 10^{-9} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 5 & 3 & -6 \end{vmatrix}$$

### Expanding the Determinant

Now, we carefully expand the determinant along the first row:

For the $\hat{i}$ component: $(3)(-6) - (4)(3) = -18 - 12 = -30$

For the $\hat{j}$ component (remember the negative sign!): $-[(2)(-6) - (4)(5)] = -[-12 - 20] = -(-32) = +32$

For the $\hat{k}$ component: $(2)(3) - (3)(5) = 6 - 15 = -9$

Putting it all together, we get the net force vector:

$$\mathbf{F}_{	ext{net}} = 10^{-9} (-30\hat{i} + 32\hat{j} - 9\hat{k}) 	ext{ N}$$

The problem states that the force is $\mathbf{F} 	imes 10^{-9} 	ext{ N}$. By comparing our result with this format, we can clearly see that the vector $\mathbf{F}$ is:

**$$\mathbf{F} = -30\hat{i} + 32\hat{j} - 9\hat{k}$$**

This matches option (b) perfectly. As a fun exercise, try taking the dot product of this force vector with the original velocity vector. You will find it equals exactly zero, confirming our physical intuition!

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Visualized Solution

$Visualizing the Vectors$

$Lorentz Force Formula$

$Setting up the Cross Product$

$Simplifying the Determinant$

$Expanding the Determinant$

$Final Force Vector$

$Physical Significance$

The Sigma Insight: Motion of a Charge in Magnetic Fields

Analyzing the Setup

The Master Equation

Setting up the Calculation

Expanding the Determinant

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Visualized Solution

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

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

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

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

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

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

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