In a particular region, there is a uniform magnetic field with a magnitude of 2.

Question

In a particular region, there is a uniform magnetic field with a magnitude of 2.00 T. You take a particle with a charge of +6.00 10-6 C, and give it an initial speed of 2.80 105 m/s in this field. Assume that, after you release it, the particle is influenced only by the magnetic field. (a) Under these conditions, what is the maximum possible magnitude of the magnetic force experienced by the particle? N (b) Under these conditions, what is the mimimum possible magnitude of the magnetic force experienced by the particle? N (c) If the particle experiences a force with a magnitude that is 25.0% of the magnitude of the maximum force, what is the angle between the particle's velocity and the magnetic field? Use an angle between 0° and 90°.

Explanation / Answer

B = 2.0 T
q = 6.0 *10^-6 C
v = 2.8 * 10^5 m/s

Now, Magetic Force on a charged moving particle is given by,
F = q* vXB
F = q* v*B* sin()

(a)
For maximum possible magnitude,
sin() = 1
F = q* v*B
F = 6.0 *10^-6 *  2.8 * 10^5 * 2.0
F = 3.36 N

(b)
For minimum possible magnitude,
sin() = 0
F = 0 N

(c)
Now F = 0.25*Fmax

0.25*Fmax = q* v*B* sin()
0.25*Fmax = Fmax * sin()
sin() = 0.25
= 14.5o

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