1. A positively charged particle of mass 7.70 × 10 -8 kg is traveling due east w
ID: 1872995 • Letter: 1
Question
1. A positively charged particle of mass 7.70 × 10-8 kg is traveling due east with a speed of 86.0 m/s and enters a 0.273-T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 3.78 × 10-3 s, at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge.
2. In a certain region, the earth's magnetic field has a magnitude of 5.5 × 10-5 T and is directed north at an angle of 46° below the horizontal. An electrically charged bullet is fired north and 19° above the horizontal, with a speed of 530 m/s. The magnetic force on the bullet is 3.5 × 10-10 N, directed due east. Determine the bullet's electric charge, including its algebraic sign (+ or -).
Explanation / Answer
1) m = 7.7*10^-8 kg
v = 86 m/s
B = 0.273 T
T/4 = 3.78*10^-3 s
==> T = 3.78*10^-3*4
= 0.01512 s
we know, T = 2*pi*m/(B*q)
q = 2*pi*m/(B*T)
= 2*pi*7.7*10^-8/(0.273*0.01512)
= 1.17*10^-4 C or 117 micro C <<<<<<------Answer for part b)
magnitude of magnetic force,
F = q*v*B*sin(theta)
= 1.17*10^-4*86*0.273*sin(90)
= 0.00274 N <<<<<<------Answer for part a)
2) B = 5.5*10^-5 T
v = 530 m/s
F = 3.5*10^-5 N
the angle between v and B = 46 + 19
= 65 degrees
we know, F = q*v*B*sin(theta)
q = F/(v*B*sin(theta))
= 3.5*10^-5/(530*5.5*10^-5*sin(65))
= 1.32*10^-3 C
sign : "-"
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