1. An alpha particle (two protons and two neutrons) traveling east at 2.0 x 105 m/s enters a magnetic field of 0.20 T pointing straight up. What is the force acting on the alpha particle?

answer:

With the fingers of the right hand pointing straight up, and the thumb pointing east, the palm points south.

F = qvBsinø = (2 x 1.6 x 10-19 C)(2.0 x 105 m/s)(0.20 T)sin90º = 1.28 x 10-14 N [S].

2. An electron traveling to the left, moves into a magnetic field directed toward the observer. Trace the path of the particle, assuming it eventually leaves the field.

An electron traveling to the left, moves into a magnetic field directed toward the observer.

answer:

The moment the electron enters the field, it experiences a force perpendicular to its velocity. The electron follows a circular path until it leaves the field.

The moment the electron enters the field, it experiences a force perpendicular to its velocity.

3. A horizontal conductor is carrying 5.0 A of current to the east. A magnetic field of 0.20 T pointing straight up cuts across 1.5 m of the conductor. What is the force acting on the conductor?

answer:

With the fingers of the right hand pointing straight up, and the thumb pointing east, the palm points south.

F = BILsinø = (0.20 T)(5.0 A)(1.5 m)sin90º = 1.5 N [S].

4. A 50.0 cm horizontal section of conductor with a mass of 8.00 g is in a 0.400 T magnetic field directed to the west. What are the magnitude and direction of current required to make this section of the conductor seem weightless?

answer:

The magnetic force must be opposite and equal to the weight of the section of the conductor.

With the fingers of the right hand pointing west, and the palm facing straight up, the thumb points north.

The weight of the conductor is mg = (0.00800 kg)(9.8 N/kg) = 0.0784 N .

The magnetic force on the conductor is

F = BILsinø , so

0.0784 N = (0.400 T)I(0.50 m)sin90º

I = 0.392 A [N]

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