A uniform magnetic field, B= ?20 mT Z^ acts on a -5.0nC charge with mass 1.0g tr

Question

A uniform magnetic field, B= ?20mT Z^ acts on a -5.0nC charge with mass 1.0g travelling with velocity v=(-2,2,1)m/s

a) What is the component of the velocity in the direction parallel to the magnetic field?

b) What is the component of the velocity perpendicular to the magnetic field?

c) Calculate the radius r of the trajectory.

d) The electron will travel in a spiral (also called a helix).Why? What is the direction of axis of the spiral? How far along the axis will the electron progress in the time it takes to make one loop of the spiral? (In other words, what is the pitch, p, of the spiral?)

Explanation / Answer

here,

B = - 20 mT k

B = - 0.02 k T

charge , q = - 5 nC

q = - 5 * 10^-9 C

mass , m = 1g

m = 1* 10^-3 kg

velocity , v = (-2 i + 2 j + k) m/s

(a)

the component of the velocity in the direction parallel to the magnetic field , V1 = B.v / |B|

V1 = 0.02 / 0.02

V1 = 1 K m/s

the component of the velocity in the direction parallel to the magnetic field is -1 k m/s

(b)

the component of the velocity perpendicular to the magnetic field , v2 = v - v1

v2 = (-2 i + 2 j ) m/s

the component of the velocity perpendicular to the magnetic field is (-2 i + 2 j ) m/s

(c)

|v2| = sqrt(2^2 + 2^2 )

|v2| = 2.83 m/s

radius of trajectory , r = m*v /(q*B)

r = 0.001 * 2.83 /(5 * 10^-9 * 0.02)

r = 2.83 * 10^7 m

radius of trajectory is 2.83 * 10^7 m

(d)

time period , T = 2 * pi* m/(q * B)

T = 2 * pi* 0.001/(5 * 10^-9 * 0.02)

T = 6.28 * 10^7 s

the time it takes to make one loop of the spiral is 6.28 * 10^7 s

the pitch , p = v1 * T

p = 1 * 6.28 * 10^7

p = 6.28 * 10^7 m

pitch of the spiral is 6.28 * 10^7 m

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