A potential difference of 600 V is applied across the plates of a parallel plate condenser. The sepa

Q: Why does the electric force point to the left if the electric field points to the right?

The electric force on a charge is given by $\mathbf{F} = q\mathbf{E}$. Since an electron has a negative charge ($q = -e$), the force vector is in the exact opposite direction of the electric field vector.

Q: Why do we completely ignore gravity in this problem?

The mass of an electron is incredibly small ($9.1 \times 10^{-31} \text{ kg}$). The gravitational force ($mg$) is many orders of magnitude weaker than the electric and magnetic forces involved here, so it has no measurable effect on the electron's path.

Q: How exactly do I use the right-hand rule here?

Point the fingers of your right hand in the direction of velocity $\mathbf{v}$ (upwards). You want the magnetic force $\mathbf{F}_m$ to be to the right. Since $\mathbf{F}_m = -e(\mathbf{v} \times \mathbf{B})$, the cross product $\mathbf{v} \times \mathbf{B}$ must point to the left. Curl your fingers so your thumb points left. Your fingers are now curling into the page, which is the direction of $\mathbf{B}$.

Q: What if the particle was a proton instead of an electron?

A proton has a positive charge. The electric force would point to the right (same as $\mathbf{E}$). To cancel it, the magnetic force must point to the left. Since $q$ is positive, $\mathbf{v} \times \mathbf{B}$ must also point to the left. Using the right-hand rule, $\mathbf{B}$ would still need to point into the page!

Q: Why does the solution mention this is the 'minimum' magnetic field?

The magnetic force is $evB \sin\theta$. We assumed $\theta = 90^\circ$ (velocity perpendicular to the magnetic field) to get $B = E/v$. If the field was at an angle, $\sin\theta$ would be less than 1, meaning we would need a larger $B$ to achieve the same force magnitude. Thus, $0.1 \text{ T}$ is the minimum required field.

Question

A potential difference of $600 	ext{ V}$ is applied across the plates of a parallel plate condenser. The separation between the plates is $3 	ext{ mm}$. An electron projected vertically, parallel to the plates, with a velocity of $2 	imes 10^6 	ext{ m/s}$ moves undeflected between the plates. Find the magnitude and direction of the magnetic field in the region between the condenser plates. (Neglect the edge effects).
(Charge of the electron = $1.6 	imes 10^{-19} 	ext{ C}$)

Sigma Percentile · Accepted Answer

### The Velocity Selector: A Delicate Balance

Imagine you are an electron, zooming vertically upwards at a blistering speed of $2 	imes 10^6 	ext{ m/s}$. Suddenly, you enter a region between two parallel plates. The plate on your left is positively charged, and the one on your right is negatively charged, creating a potential difference of $600 	ext{ V}$ across a tiny $3 	ext{ mm}$ gap.

This setup creates a strong electric field $\mathbf{E}$ pointing from the positive to the negative plate (left to right). Because you are negatively charged, you immediately feel a powerful electric force $\mathbf{F}_e$ pulling you towards the positive plate on the left. If nothing else happens, you will crash into it!

### The Master Equation for Undeflected Motion

But the problem states you move **undeflected**. This means there must be another force perfectly counteracting the electric force. Enter the magnetic field $\mathbf{B}$. It exerts a magnetic force $\mathbf{F}_m$ on you. For you to travel in a straight line, the net force must be zero:

$$\mathbf{F}_{	ext{net}} = \mathbf{F}_e + \mathbf{F}_m = 0$$

This implies that the magnetic force must be exactly equal in magnitude and opposite in direction to the electric force.

$$|\mathbf{F}_e| = |\mathbf{F}_m|$$

We know the electric force is $eE$ and the magnetic force is $evB \sin	heta$. Assuming the magnetic field is perpendicular to your velocity (which gives the maximum force for a given field strength, hence the "minimum" required field), $\sin(90^\circ) = 1$.

$$eE = evB$$

Solving for the magnetic field $B$, we get:

$$B = \frac{E}{v}$$

Since the electric field $E$ in a parallel plate capacitor is the voltage $V$ divided by the distance $d$ ($E = \frac{V}{d}$), we can substitute this in:

$$B = \frac{V}{dv}$$

### Final Calculation and Direction

Now, let's plug in the numbers. We have $V = 600 	ext{ V}$, $d = 3 	imes 10^{-3} 	ext{ m}$, and $v = 2 	imes 10^6 	ext{ m/s}$.

$$B = \frac{600}{(3 	imes 10^{-3}) 	imes (2 	imes 10^6)}$$

$$B = \frac{600}{6 	imes 10^3} = 0.1 	ext{ T}$$

So, the magnitude of the magnetic field is **$0.1 	ext{ T}$**.

What about the direction? The electric force $\mathbf{F}_e$ is pulling you to the left ($-x$ direction). Therefore, the magnetic force $\mathbf{F}_m$ must push you to the right ($+x$ direction).

The Lorentz force law states $\mathbf{F}_m = q(\mathbf{v} 	imes \mathbf{B})$. Since your charge $q$ is negative ($-e$), the vector $\mathbf{v} 	imes \mathbf{B}$ must point in the opposite direction of the force, which is the $-x$ direction.

Using the right-hand rule: point your fingers in the direction of your velocity (upwards, $+y$), and you need your thumb (representing $\mathbf{v} 	imes \mathbf{B}$) to point left ($-x$). To do this, your palm must face into the page. Therefore, the magnetic field $\mathbf{B}$ must point **perpendicularly into the paper** ($-z$ direction).

Visualized Solution

$S e t u p o f t h e C o n d e n ser$

$E l ec t r i c F i e l d an d F or ce$

$C o n d i t i o n f or U n d e f l ec t e d M o t i o n$

$E q u a t in g F or ce M a g ni t u d es$

$C a l c u l a t in g M a g n e t i c F i e l d M a g ni t u d e$

$D e t er minin g t h e D i r ec t i o n o f B$

$W h y M inim u m M a g n e t i c F i e l d ?$

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Velocity Selector

The Master Equation for Undeflected Motion

Final Calculation and Direction

Similar Questions

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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

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

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Undeflected Motion

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Determining the Direction of B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Why Minimum Magnetic Field?