Crossed electric and magnetic fields. A charged particle...

Crossed electric and magnetic fields. A charged particle moves in the $x$ direction through a region in which there is an electric field $E_{y}$ and a perpendicular magnetic field $B_{z}$. What is the condition necessary to ensure that the net force on the particle will be zero? Show the $\mathbf{v}, \mathbf{E}$, and $\mathbf{B}$ vectors on a diagram. What is the condition on $v_{x}$ if $E_{y}=10$ statvolts $/ \mathrm{cm}$ and $B_{z}=300 \mathrm{G}$ ? $\quad$ Ans. $v_{x}=1 \times 10^{9} \mathrm{~cm} / \mathrm{s}$

Key Concepts

Lorentz Force Law

This concept states that a charged particle moving in electric and magnetic fields experiences a force given by F = q(E + v × B). The idea is that the electromagnetic force is a combination of the electric force, which is independent of the particle’s velocity, and the magnetic force, which depends on the cross product of the velocity and the magnetic field. Balancing these forces can lead to a condition where the net force is zero, allowing the particle to pass through without deflection.

Velocity Selector in Crossed Fields

A velocity selector uses perpendicular electric and magnetic fields so that only particles with a specific velocity, for which the electric force perfectly cancels the magnetic force, travel undeflected. The cancellation condition is derived from setting qE equal to q(v × B), leading to the relation between the particle’s speed and the field strengths. This principle is widely used in experiments and devices to filter charged particles by their speeds.

Electromagnetic Unit Systems (CGS/Gaussian)

In electromagnetic problems, especially in the context of the CGS (Gaussian) unit system, the equations of motion for charged particles include factors such as the speed of light when relating electric and magnetic forces. For example, when calculating the equilibrium condition where the electric and magnetic forces cancel, the magnetic force term includes a factor of 1/c, making the velocity condition v = c(E/B) in CGS units. Understanding these unit-specific factors is crucial for correctly applying physical laws in different measurement systems.

Transcript

00:01 Hello everyone welcome back so today we are going to solve this problem which states crossed the electric field and magnetic field the charged particle moves in the x direction through a region which there is an electric field ey and perpendicular magnetic field will be read so what is the necessary condition to ensure that the net course on the particle will be zero so that that v and v3 is on the i run and what is the condition on vx if by is e per sotene x if b y is a plus to 10 steadfolds per centimeter and bz is equals to 300 bs so we need to find the condition we know that by lawrence force and a charge particle moves and then like even then medically so we post units few times fifted plus v -velect cross t vector what we could be this problem if we drop the partition coordinate system then we like this so the cartesian coordinate system x bide and c cap now, magnetic field in this direction and electric wheel is in the direction and particle is thrown in next up direction.

02:02 This is the direction of particle in which it is thrown.

02:08 So as given in the version of the net force, we need convert 0, then q times e vector v cross v vector so from here we got v vector is equal to minus time b cross b since we didn't know the vector value of b vector so let the value of speed b in n tap direction and the magnetic will be given as 300 goss and 1 cos the reverse to 10 minus 4 tesla in the direction of see cap and direction of electric field in my direction which is 10 stat volts and 1 set volt to 300 volts multiply 3 over and 1 centimeter is it is to 10 restore minus 2 meter so this hole will come out to get a distor 2 and this in volt per meter so as we have converted all the values in the size system so this will make our life easier now now, let's put the values here.

04:05 So if we put the value of electric field, we get three times 10 list over 6 in the direction by gap.

04:22 And we get minus times v cap cross v cap.

04:27 So v is in arbitratory n -cap direction, which we don't move where it is.

04:34 So we write v.

04:35 Times n -cap value of validity field 0 .03 times z -kept.

04:47 So from vector we know that if i -capped cross z -caf then we get direction of minus or bichet so here n -keh should be equals to i -cap to cat to cat the equal direction right here because the negative sign is multiple here...

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