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**1. 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 [Broken]

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?

**2. 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]

**3. 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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