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Homework Statement
Problem 33.74
The switch in the figure has been open for a long time. It is closed at t = 0s.
PART A - What is the current through the 20[tex]\Omega[/tex] resistor immediately after the switch is closed?
PART B - What is the current through the 20[tex]\Omega[/tex] resistor immediately after the switch has been closed for a long time?
PART C - What is the current through the 20[tex]\Omega[/tex] resistor immediately after the switch is reopened?
Problem 32.62
An electron travels with speed 0.900 x 10^7 m/s between the two parallel charged plates shown in the figure. The plates are separated by 1.0 cm and are charged by a 200 V battery.
What magnetic field strength (and direction) will allow the electron to pass between the plates without being deflected
Problem 32.66
The uniform 30.0 mT magnetic field in the figure points in the positive z-direction. An electron enters the region of magnetic field with a speed of 5.30 x 10^6 m/s and at an angle of 30 degrees above the xy-plane
http://session.masteringphysics.com/myct/problemWork?template=problemView&assignmentProblemID=2728400
PART A - Find the radius (in mm) of the electron's spiral trajectory.
PART B - Find the pitch of the electron's spiral trajectory.
Problem 32.74
A bar magnet experiences a torque of magnitude 0.075 Nm when it is perpendicular to a 0.50 T external magnetic field. What is the strength of the bar magnet's on-axis magnetic field at a point 20 cm from the center of the magnet?
Homework Equations
I = V/R
[tex]\Delta V = w v_d B[/tex]
[tex]r = \frac{mv}{qB}[/tex]
[tex]\tau = \vec{\mu} B sin\vartheta[/tex]
[tex]B = \frac{\mu_0}{4 \pi} \frac{2 \vec{\mu}}{z^3}[/tex]
The Attempt at a Solution
Problem 33.74
PART A
I = V/R = 30/20 = 3/2 A
PART B
Same has part A
PART C
Since there is an inductor in the circuit, the current would still be there if the switch has just been opened - so same as the other two parts.
Problem 32.62
[tex]\Delta V = w v_d B \Rightarrow B = \frac{\Delta V}{w v_d} = \frac{200}{(0.01)(0.9*10^7)} = 2.22*10^-4[/tex]
For the direction, using the right hand rule that related magnetic field and speed; since the index finger points to the right, the middle finger (which is perpindicular to the index finger) is pointing out of the page.
Problem 32.66
PART A
[tex]r = \frac{mv}{qB} = \frac{(9.11*10^-31)(5.3*10^6)}{(1.6*10^-19)(30*10^-3)} = 10^-3[/tex]
PART B
I've got no idea. I can't even find pitch in my textbook.
Problem 32.74
since its perpendicular, sin90 = 1 so:
[tex]\tau = \vec{\mu} B \Rightarrow \vec{\mu} = \frac{\tau}{B} = \frac{0.075}{0.5} = 0.15[/tex]
[tex]B = \frac{\mu_0}{4 \pi} \frac{2 \vec{\mu}}{z^3} = 10^-7 \frac{2(0.15)}{0.2^3} = 37.5 T[/tex]
Any help would be appreciated. Thank You.
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