A thin, horizontal copper rod is 1.18 m long and has a mass of 60.6 g . What is

Question

A thin, horizontal copper rod is 1.18 m long and has a mass of 60.6g. What is the minimum current in the rod that can cause it to float in a horizontal magnetic field of 2.29 T?

A charged particle is moving perpendicularly to a magnetic field B. Fill in the blank indicating the direction for the quantity missing in each option.

Use the diagram above for the directions of the various axes.
+x-x+y-y+z-z Negative Charge, Velocity: ???, B-Field: -z, Force: +y
+x-x+y-y+z-z Positive Charge, Velocity: +y, B-Field: ???, Force: -z
+x-x+y-y+z-z Positive Charge, Velocity: +y, B-Field: -z, Force: ???

A charged particle moves through a velocity selector at a constant speed in a straight line. The electric field of the velocity selector is 3.61E+3 N/C, while the magnetic field is 0.397 T. When the electric field is turned off, the charged particle travels on a circular path whose radius is 4.03 cm. Calculate the charge-to-mass ratio of the particle.

The solar wind is a thin, hot gas given off by the sun. Charged particles in this gas enter the magnetic field of the earth and can experience a magnetic force. Suppose a charged particle traveling with a speed of 9.08E+6 m/s encounters the earth's magnetic field at an altitude where the field has a magnitude of 1.30E-7 T. Assuming that the particle's velocity is perpendicular to the magnetic field, calculate the radius of the circular path on which the particle would move if it were an electron.

A thin, horizontal copper rod is 1.18 m long and has a mass of 60.6g. What is the minimum current in the rod that can cause it to float in a horizontal magnetic field of 2.29 T? A charged particle is moving perpendicularly to a magnetic field B. Fill in the blank indicating the direction for the quantity missing in each option. Use the diagram above for the directions of the various axes. A charged particle moves through a velocity selector at a constant speed in a straight line. The electric field of the velocity selector is 3.61E+3 N/C, while the magnetic field is 0.397 T. When the electric field is turned off, the charged particle travels on a circular path whose radius is 4.03 cm. Calculate the charge-to-mass ratio of the particle. The solar wind is a thin, hot gas given off by the sun. Charged particles in this gas enter the magnetic field of the earth and can experience a magnetic force. Suppose a charged particle traveling with a speed of 9.08E+6 m/s encounters the earth's magnetic field at an altitude where the field has a magnitude of 1.30E-7 T. Assuming that the particle's velocity is perpendicular to the magnetic field, calculate the radius of the circular path on which the particle would move if it were an electron.

Explanation / Answer

1.

F = q(v * B)

Fnety = ma = 0 = F - Fg
F = Fg
q(v * B) = mg

I = q/t
q = I * t

(I * t)(x/t * B) = mg
(I * x * B) = mg
I = mg/xB
I = (.0606)(9.8)/(1.18)(2.29)
I = 0.219 A

2.

use right hand thumb rule

1. v = +y
2. b = -y,
3. -y

3.

q/m =E/(B^2*r) = (3610)/(.397^2*.0403)= 568356.9

4.

Bqv = mv^2/r
For electron:
r = mv/Bq = v / [B(q/m)] =9.08e+6/ (1.3e-7*1.76e+11)
=9.08e+6/2.288e+4 = 396.853 m
For proton
r = r electron / 1836 = 2971m

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