A particle of charge -16 x 10^-18 C moving with velocity 10 ms^-1 along the X-axis enters a region w

Q: Why does the electric force point along the positive z-axis when the electric field is along the negative z-axis?

The electric force on a charge is given by $\mathbf{F} = q\mathbf{E}$. If the charge $q$ is positive, the force is in the same direction as the field. But if the charge is negative (like in this problem), the negative sign flips the direction of the force to be exactly opposite to the electric field.

Q: How do we find the direction of the magnetic force?

We use the right-hand rule for the cross product $\mathbf{v} \times \mathbf{B}$. Point your fingers in the direction of velocity (x-axis) and curl them towards the magnetic field (y-axis). Your thumb points in the positive z-axis. However, because the formula is $\mathbf{F} = q(\mathbf{v} \times \mathbf{B})$ and $q$ is negative, you must flip the final result. Thus, the magnetic force points along the negative z-axis.

Q: Why do we set the net force to zero?

The problem states that the particle 'continues moving along the X-axis'. This implies it does not accelerate in the y or z directions. According to Newton's First Law, if there is no change in velocity (no deflection), the net force acting on the particle must be zero.

Q: What is a Weber per square meter (Wb/m²)?

Weber per square meter ($\text{Wb/m}^2$) is the SI unit for magnetic field (or magnetic flux density). It is exactly equivalent to one Tesla (T).

Q: Does the mass of the particle matter in this problem?

No, the mass does not matter here. Because the net force is zero, there is no acceleration. The mass would only be relevant if the forces were unbalanced and we needed to calculate the resulting acceleration or the radius of a curved path.

Question

A particle of charge $-16 	imes 10^{-18}$ C moving with velocity $10 	ext{ ms}^{-1}$ along the X-axis enters a region where a magnetic field of induction $B$ is along the Y-axis and an electric field of magnitude $10^4 	ext{ Vm}^{-1}$ is along the negative Z-axis. If the charged particle continues moving along the X-axis, the magnitude of $B$ is

Options:

(A) $10^3 	ext{ Wb/m}^2$

(B) $10^5 	ext{ Wb/m}^2$

(C) $10^{16} 	ext{ Wb/m}^2$

(D) $10^{-3} 	ext{ Wb/m}^2$

Sigma Percentile · Accepted Answer

This problem is a classic demonstration of the **Lorentz Force** and the principle behind a **Velocity Selector**. When a charged particle moves through a region containing both electric and magnetic fields, it experiences forces from both. Let's break down how these forces interact to allow the particle to move completely undeflected.

### Analyzing the Setup

Imagine a 3D coordinate system. We are given a particle with a negative charge, $q = -16 	imes 10^{-18}$ C, moving along the positive x-axis with a velocity $\mathbf{v} = 10 \hat{i} 	ext{ ms}^{-1}$.

It enters a region where two fields exist simultaneously:
1. A magnetic field $\mathbf{B}$ pointing along the y-axis: $\mathbf{B} = B \hat{j}$
2. An electric field $\mathbf{E}$ pointing along the negative z-axis: $\mathbf{E} = -10^4 \hat{k} 	ext{ Vm}^{-1}$

The problem states that the particle continues moving along the x-axis without any deflection. For this to happen, the net force acting on the particle must be exactly zero. This means the electric force and the magnetic force must perfectly balance each other out.

### The Electric Force

The electric force $\mathbf{F}_e$ acting on a charge is simply the product of the charge and the electric field:

$$\mathbf{F}_e = q\mathbf{E}$$

Substituting our values:

$$\mathbf{F}_e = (-16 	imes 10^{-18} 	ext{ C})(-10^4 \hat{k} 	ext{ V/m})$$

Notice that multiplying two negative values gives a positive result. This makes physical sense: a negative charge experiences an electric force in the direction *opposite* to the electric field. Since the field is along the negative z-axis, the force is along the positive z-axis.

$$\mathbf{F}_e = 16 	imes 10^{-14} \hat{k} 	ext{ N}$$

### The Magnetic Force

Now, let's calculate the magnetic force $\mathbf{F}_m$. The magnetic force on a moving charge is given by the cross product of its velocity and the magnetic field:

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

Substituting our vectors:

$$\mathbf{F}_m = (-16 	imes 10^{-18})(10 \hat{i} 	imes B \hat{j})$$

Using the right-hand rule for cross products, we know that $\hat{i} 	imes \hat{j} = \hat{k}$. However, because our charge $q$ is negative, the final direction of the force flips to the negative z-axis.

$$\mathbf{F}_m = -16 	imes 10^{-18} 	imes 10 B \hat{k}$$

$$\mathbf{F}_m = -16 	imes 10^{-17} B \hat{k} 	ext{ N}$$

### Balancing the Forces

For the particle to remain undeflected, the sum of the electric and magnetic forces must be zero:

$$\mathbf{F}_e + \mathbf{F}_m = 0$$

$$16 	imes 10^{-14} \hat{k} - 16 	imes 10^{-17} B \hat{k} = 0$$

We can equate the magnitudes of the two forces:

$$16 	imes 10^{-14} = 16 	imes 10^{-17} B$$

Now, we simply solve for the unknown magnetic field magnitude $B$:

$$B = \frac{16 	imes 10^{-14}}{16 	imes 10^{-17}}$$

The $16$s cancel out beautifully, and using the laws of exponents, we subtract the denominator's exponent from the numerator's:

$$B = 10^{-14 - (-17)} = 10^3 	ext{ Wb/m}^2$$

**The required magnetic field is $10^3 	ext{ Wb/m}^2$.** This delicate balance of forces is exactly how a velocity selector in a mass spectrometer works, ensuring only particles with a specific speed $v = E/B$ pass through!

Select Answer:

Visualized Solution

Visualizing the Setup

Field Orientations

Condition for Undeflected Motion

Electric Force Setup

Calculating Electric Force

Magnetic Force Setup

Calculating Magnetic Force

Equating the Forces

Final Calculation

The Velocity Selector

The Sigma Insight: Motion of a Charge in Magnetic Fields

Analyzing the Setup

The Electric Force

The Magnetic Force

Balancing the Forces

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