Charged particle in a magnetic field

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

Velocity of particle at any time t is

v = v_xi + v_yj

Net Force on the particle F = (2qv_yB)i + (2qv_xB)j

Net acceleration of the particle = F/m

Magnitude of net acceleration = (2qvB)/m

Particle has an initial velocity as well as acceleration along an angle 45° with +x direction which means it moves along the straight line x = y .

I wonder why the question talks about finding the center of the circular path followed by the particle .

I might be making mistake .

 

[Chandra Prayaga]

Your calculation of the net force is wrong. Do the cross product carefully.

 

[Jahnavi]

Sorry .

Net Force on the particle F = (2qv_yB)i -(2qv_xB)j

If I solve the two eq of motion in x and y directions , I end up with v_x² + v_y² =v² .

I might be making mistake again .

 

[Chandra Prayaga]

What is the path with this correct force?

 

[Jahnavi]

How do I do that ? I solved the two eqs of motion and ended up with the above result .

 

[Chandra Prayaga]

There are a couple of ways of doing this depending on the level at which you are working. Do you know how to solve a differential equation of motion? For example, did you solve the differential equation for a simple harmonic oscillator, or for motion with constant acceleration?

 

[Jahnavi]

mdv_x/dt = 2qv_yB

mdv_y/dt = -2qv_xB

Solving them gives v_x² + v_y² = v² which means speed of the particle is constant .

How do I find the path ?

 

[Chandra Prayaga]

How did you solve those two differential equations?

 

[Jahnavi]

Dividing first eq by second and integrating both sides .

Limits of v_x is v/√2 to v_x

Limits of v_y is v/√2 to v_y

∫v_xdv_x = -∫v_ydv_y

 

[Chandra Prayaga]

Try solving them like this:
1. Differentiate each equation with respect to t once more. The first equation will then have dv_y/dt in it. For that substitute from the first equation. This will give you a familiar equation in v_x, without the v_y in it. Similalry get an equation for v_y alone. This technique will "decouple" the equations for the two variables v_x and v_y. You can then solve each separately, given the initial conditions.

 

[Chandra Prayaga]

Sorry. There was a typo. The third sentence in item 1 should have been, "For that substitute from the second equation."

 

[Jahnavi]

v_x = vsin(ωt+π/4) , ω=2qB/m

v_y = -vsin(ωt-π/4)

Is that correct ?

 

[Chandra Prayaga]

Do these agree with your initial conditions?

 

[Jahnavi]

Yes .

 

[Chandra Prayaga]

Now, since v_x = dx/dt, and v_y = dy/dt, you have two 1st order differential equations to solve, again with given initial conditions. Go ahead and solve them.

 

[Jahnavi]

Solving further gives ,

x = v/(√2ω) - (v/ω)cos(ωt+π/4)

y= -v/(√2ω) + (v/ω)cos(ωt-π/4)

 

[Chandra Prayaga]

Well, now you have x and y as functions of time, so you have the actual path. Show that it is a circle, and you will also get the radius.

 

[Jahnavi]

Squaring and adding is quite messy .

It looks difficult to manipulate terms . There are cos²(A+B) , cos²(A-B) terms .

Algebra looks far more difficult , rather terrible :)

 

[Chandra Prayaga]

Try taking the constant terms to the left hand side and then square and add.

 

[Jahnavi]

The trick worked

Center is [ (mv)/(2√2qB) , (-mv)/(2√2qB) , 0 ]

Is that correct ?

 

[Chandra Prayaga]

Certainly looks like that, unless something went wrong in the arithmetic. You can check that for yourself. You should draw a diagram, and make sure that the circle looks correct. For example, when you projected the particle from the origin, in which direction did it start? What was the dorection of the force? Which way would the force bend the path of the particle? Form that you can predict which way the circle would look, and where you can expect the center. That is, are you expecting the circle to bend upwards, or downwards? Does that agree with your final answer? All such questions will become clear if you draw the diagram and the initial velocity and force vectors.

Good luck!

 

[Jahnavi]

Thank you so much :)