Exercise 27.8 A particle with charge 6.00 nC is moving in a uniform magnetic fie

Question

Exercise 27.8

A particle with charge 6.00 nC is moving in a uniform magnetic field B =( 1.21 T )k^. The magnetic force on the particle is measured to be F =( 3.60×107 N )i^+( 7.60×107 N )j^.

Part A

Are there components of the velocity that are not determined by the measurement of the force?

Part B

Calculate the x-component of the velocity of the particle.

Part C

Calculate the y-component of the velocity of the particle.

Part D

Calculate the scalar product v F .

Part E

What is the angle between v  and F ? Give your answer in degrees.

Explanation / Answer

A)

  Measuring force alone doesn't give you any information about velocity

B&C)

F = -3.6*10^-7i , 7.6*10^-7 j , 0k N
B = 0i , 0j , -1.21k T
q = -6.0*10^-9 C

Fx = (v2 * b3) - (v3 * b2) = -3.6*10^-7i / q
Fy = (v3 * b1) - (v1 * b3) = 7.6*10^-7 j / q
Fz = (v1 * b2) - (v2 * b1) = 0 k / q

Fx = (v2 * -1.3) - (v3 * 0) = -3.6*10^-7i / q
Fy = (v3 * 0) - (v1 * -1.21) = 7.6*10^-7 j / q
Fz = (v1 * 0) - (v2 * 0) = 0 k / q

vx = (7.6 * 10 ^ -7 / q) / 1.21 = 153.2666 m/s
vy = (-3.6 * 10 ^ -7 / q) / -1.21= - 72.6 m/s
vz = 0 k

D)
Now (vector V) dot (Vector F) = (Vx * Fx) + (Vy * Fy)
(since Vz = 0 and Fz = 0)
= (153.266) * (- 3.60) * 10^(- 7) + (- 72.6) * (7.60) * 10^(- 7)
= (- 551.7576 – 551.76)*10^(- 7) W
= - 11.0351 10^(- 5) W

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