1. A 30.0-g metal ball having net charge Q = 5.00 µC is thrown out of a window h

Question

1. A 30.0-g metal ball having net charge Q = 5.00 µC is thrown out of a window
horizontally at a speed v = 20.0 m/s. The window is at a height h = 20.0 m above the ground. A uniform horizontal magnetic field of magnitude B = 0.010 0 T is perpendicular to the plane of the ball’s trajectory. Find the magnetic force acting on the ball just before it hits the ground.

2. An electron is projected into a uniform magnetic field B = (1.40i + 2.10j) T. Find the vector expression for the force on the electron when its velocity is v = 3.70 × 105 j m/s.

3. Imagine a very long, uniform wire with a linear mass density of 1.00 g/m that encircles the Earth at its magnetic equator. Suppose that the planet’s magnetic field is 50.0 µT horizontally north throughout this region. What are the magnitude and direction of the current in the wire that keep it levitated just above the ground?

4. A section of conductor 0.400 cm thick is used in a Hall-effect measurement. A Hall voltage of 35.0 µV is measured for a current of 21.0 A in a magnetic field of 1.80 T. Calculate the Hall coefficient for the conductor.

Explanation / Answer

Since you are only allowed to ask one question per post in Cramster, I can only answer one. I will answer the forst one. Please re-post the others separately for solutions.

Question 1

The magnetic force is found by qvB, but the velocity and the magnetic fields must be perpendicular to each other. The 20 m/s can not be used, it is the horizontal velocity, and with a horizontal magnetic field, it will not solve the problem.

We need the y velocity it will have just prior to striking the ground.

Since the charge is 20 m high, and has zero initial y velocity, we can use

vf2 = vo2 + 2ad to find the velocity

vf = [(2)(9.8)(20)]

vf = 19.8 m/s

Then, F = qvB

F = (5 X 10-6)(19.8)(.01)

F = 9.9 X 10-7 N

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