An electron gun G emits electrons of energy 2 keV travelling in the positive x-direction. The electr

Q: Why does the electron move in a helix?

Because its velocity has an angle with the magnetic field. The perpendicular component creates a circular motion, while the parallel component pulls it forward, creating a 3D spiral or helix.

Q: Why did we take $n=1$ for the minimum magnetic field?

The pitch of the helix is inversely proportional to the magnetic field ($p \propto 1/B$). To minimize $B$, we must maximize the pitch $p$. Since $GS = n \cdot p$, the maximum possible pitch that fits in the distance $GS$ is when $n=1$.

Q: What happens to the perpendicular component of velocity?

The perpendicular component ($v \sin 60^\circ$) is responsible for the circular part of the motion. It determines the radius of the helix but doesn't affect the time it takes to complete one revolution.

Q: Why do we use $v \cos 60^\circ$ for the pitch?

The pitch is the distance traveled *along* the magnetic field axis during one full rotation. Only the component of velocity parallel to the magnetic field ($v \cos 60^\circ$) contributes to this forward motion.

Q: How did we find the velocity from kinetic energy?

We used the classical kinetic energy formula $K = \frac{1}{2}mv^2$. Remember to convert the given energy from keV to Joules by multiplying by $1.6 \times 10^{-16}$ before solving for $v$.

Question

An electron gun $G$ emits electrons of energy $2	ext{ keV}$ travelling in the positive $x$-direction. The electrons are required to hit the spot $S$ where $GS = 0.1	ext{ m}$, and the line $GS$ makes an angle of $60^\circ$ with the $x$-axis as shown in figure. A uniform magnetic field $\mathbf{B}$ parallel to $GS$ exists in the region outside the electron gun.

Find the minimum value of $B$ needed to make the electrons hit $S$.

Sigma Percentile · Accepted Answer

### The Setup: A 3D Dance of an Electron

Imagine an electron gun $G$ firing an electron into a region with a uniform magnetic field $\mathbf{B}$. The electron is shot along the $x$-axis, but the magnetic field is tilted, making an angle of $60^\circ$ with the electron's initial velocity.

When a charged particle enters a magnetic field at an angle, its velocity can be broken down into two components: one parallel to the field ($v_{\parallel}$) and one perpendicular to it ($v_{\perp}$). The perpendicular component causes the electron to move in a circle, while the parallel component pulls it forward along the magnetic field lines. The combination of these two motions creates a beautiful 3D spiral, known as a **helical path**.

### Decoding the Helix: Pitch and Time Period

Before we dive into the helix, let's find the speed of our electron. We are given its kinetic energy $K = 2 	ext{ keV}$. Using the classical kinetic energy formula $K = \frac{1}{2}mv^2$, we can solve for the velocity $v$:

$$v = \sqrt{\frac{2K}{m}}$$

Converting the energy to Joules ($1 	ext{ eV} = 1.6 	imes 10^{-19} 	ext{ J}$) and plugging in the mass of an electron ($9.1 	imes 10^{-31} 	ext{ kg}$), we get:

$$v = \sqrt{\frac{2 	imes 2 	imes 10^3 	imes 1.6 	imes 10^{-19}}{9.1 	imes 10^{-31}}} \approx 2.65 	imes 10^7 	ext{ m/s}$$

Now, let's look at the helix. The distance the electron travels along the magnetic field during one full rotation is called the **pitch** ($p$). It is simply the product of the parallel velocity and the time period of revolution ($T$):

$$p = v_{\parallel} T = (v \cos 60^\circ) \frac{2\pi m}{eB}$$

### The Target: Hitting Spot S

The problem states that the electron must hit a specific spot $S$, which lies exactly on the axis of the helix at a distance $GS = 0.1 	ext{ m}$. For the electron to hit this spot, the total distance $GS$ must be an exact integer multiple of the pitch. It could complete one, two, or many spirals before hitting $S$:

$$GS = n p \quad 	ext{where } n = 1, 2, 3, \dots$$

We are asked to find the **minimum** magnetic field $B_{	ext{min}}$. Looking at our pitch formula, the magnetic field $B$ is in the denominator. This means a smaller magnetic field results in a larger pitch. To minimize $B$, we must maximize the pitch $p$. The maximum possible pitch that fits within the distance $GS$ occurs when the electron hits $S$ after exactly one revolution, meaning $n = 1$.

### Crunching the Numbers: The Final Calculation

Setting $n = 1$, we have $GS = p$. Substituting our pitch formula, we get:

$$GS = \frac{2\pi m v \cos 60^\circ}{eB_{	ext{min}}}$$

Rearranging to solve for $B_{	ext{min}}$:

$$B_{	ext{min}} = \frac{2\pi m v \cos 60^\circ}{e (GS)}$$

Finally, we substitute all our known values into this master equation:

$$B_{	ext{min}} = \frac{2\pi (9.1 	imes 10^{-31} 	ext{ kg}) (2.65 	imes 10^7 	ext{ m/s}) (0.5)}{(1.6 	imes 10^{-19} 	ext{ C}) (0.1 	ext{ m})}$$

Calculating this yields our final answer:

$$B_{	ext{min}} = 4.73 	imes 10^{-3} 	ext{ T}$$

And there you have it! By understanding the geometry of the helical path and the relationship between the magnetic field and the pitch, we've successfully navigated this complex 3D problem.

Visualized Solution

$K = 2 keV$

$v = \frac{2 K}{m}$

$θ = 6 0^{\circ}$

$Helical Path$

$p = v_{∥} T$

$GS = n p$

$B_{min} ⟹ n = 1$

$B_{min} = \frac{2 π m v c o s 6 0 ^{\circ}}{e ( GS )}$

$B_{min} = 4.73 \times 1 0^{- 3} T$

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Setup

Decoding the Helix

The Target

Crunching the Numbers

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K=2 keV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=m2K​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ=60∘

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=v∥​T

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bmin​⟹n=1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bmin​=e(GS)2πmvcos60∘​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bmin​=4.73×10−3 T