PLEASE BE CAREFUL!, THIS IS MY 3RD TIME POSTING THIS PROBLEM, IF YOU DON\'T KNOW

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PLEASE BE CAREFUL!, THIS IS MY 3RD TIME POSTING THIS PROBLEM, IF YOU DON'T KNOW FOR SURE HOW TO DO IT, PLEASE DON'T.

A charged particle of mass m = 5.8X10-8 kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 1.3T aligned with the positive z-axis as shown. The particle enters the region at (x,y) = (0.7 m, 0) and leaves the region at (x,y) = 0, 0.7 m a time t = 480 s after it entered the region.

Please do not answer if you are not sure about any of the questions!

#4 and 5 are supposed tobe tricky, so ifyou are not entirely sure on how to do them, please do not answer at all (any of the questions), thank you.

1)

With what speed v did the particle enter the region containing the magnetic field? = m/s

2)

What is Fx, the x-component of the force on the particle at a time t1 = 160 s after it entered the region containing the magnetic field. = N

3)

What is Fy, the y-component of the force on the particle at a time t1 = 160 s after it entered the region containing the magnetic field. = N

4)

What is q, the charge of the particle? Be sure to include the correct sign. = C

5)

If the velocity of the incident charged particle were doubled, how would B have to change (keeping all other parameters constant) to keep the trajectory of the particle the same?

Increase B by a factor of 2

Increase B by less than a factor of 2

Decrease B by less than a factor of 2

Decrease B by a factor of 2

There is no change that can be made to B to keep the trajectory the same.

Explanation / Answer

here,
magnatic field, B = 1.3 T
mass of particle, m = 5.8*10^-8 kg
path radius , r = 0.7 m
time, t = 480us = 4.80*10^-4 s

Part A:

Since particle follow circular path or quater of circle,
length, L = pi*r/2 = pi*(0.7/2) = 1.1 m

So velocity, v = L/t
v = 1.1/(4.80*10^-4)
v = 2291.667 m/s

Par b:
X component of force, Fx = F*CosA
time, t = 160 us = 1.6*10^-4 s

Radians covered in time t = 160 us
A = (v/R)*t = (2291.667/0.7)*1.6*10^-4
A = 0.524 radians

Therefore,
Fx = -m*v^2/r * Cos(0.524)
Fx = (-5.8*10^-8*(2291.667)^2/0.7) * Cos(0.524)
fx = -0.377 N
-ve since force is pointing towards centre of circle,

Part 3:
Similarly,
Fy = F*SinA =- mv^2/r * SinA
Fy = -5.8*10^-8*(2291.667)^2/0.7 *Sin(0.524)
Fy = -0.2177 N

Part 4:
Since F = q*v*B
sqrt(0.377^2 + 0.2177^2) = -q*2291.667*1.3

q = -1.461*10^-4 C

-ve since field is out of page

Part 5:
If the velocity were doubled, the radius of curvature would double.
Since the radius of curvature is inversely proportional to the magnetic field strength,
if we then double the magnetic field strength , curvature will be halfed,
so

increases B by factor 2 is right answer

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