A proton having a speed of 3.0 Times 106 m/s in a direction perpendicular to a u

Question

A proton having a speed of 3.0 Times 106 m/s in a direction perpendicular to a uniform magnetic field moves in a circle of radius 0.20 m within the field. What is the magnitude of the magnetic field? (e = 1.60 Times 10-19 C, mproton m 1.67 Times 10-27 kg) 0.080 T 0.16 T 0.24 T 0.32 T 0.36 T Alpha particles, each having a charge of +2e and a mass of 6.64 Times 10-27 kg, are accelerated in a uniform 0.80-T magnetic field to a final orbit radius of 0.30 m. The field is perpendicular to the velocity of the particles. How long does it take an alpha particle to make one complete circle in the final orbit? (e = 1.60 Times 10-19 C) 0.15 mu s 0.25 mu s 0.33 mu s 0.40 mu s 0.49 mu s A Hat circular coil has 200 identical loops of very thin wire. Each loop has an area of 0.12 m2 and carries 0.50 A of current. This coil is placed in a magnetic field of 0.050 T oriented at 30 degree to the plane of the loop. What is the magnitude of the magnetic torque on the coil? 0.25 N middot m 0.52 N middot m 2.5 N middot m 5.2 N m A thin copper rod 1.0 m long has a mass of 0.050 kg and is in a magnetic field of 0.10 T. What minimum current in the rod is needed in order for the magnetic force to balance the weight of the rod? 1.2 A 2.5 A 4.9 A 9.8 A None of the above

Explanation / Answer

16. A charged particle moves through a magnetic field then it will experience a force according to
Fm = q v B sin

As it moves perpendicular, the angle is 90°, and sin90 = 1, so Fm = qVB

the charge experiencing this Force travels in a circle then there is a counter centripetal force keeping the particle on the circle given by
Fc = mv2/r
qvB = mv2/r

solving for B we get
B = mv / rq
B =1.67×1027 * 3x106 / (0.2 * 1.602×1019) = 0.16 T

19. Magnetic Force = Gravitational Force
BIL = mg
so, I = mg / BL
I = (0.05)*(9.8) / (0.1)(1)
I = 4.9 A

The other 2 questions, post them in another question thread.

Hope this helps

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