A particle with a charge of -2.5010-8 C is moving with an instantaneous velocity

Question

A particle with a charge of -2.5010-8 C is moving with an instantaneous velocity of magnitude 40.0 km/s in the xy-plane at an angle of 50.0 degree counterclockwise from the +x axis.

PartA:

What is the magnitude of the force exerted on this particle by a magnetic field with magnitude 2.00 in the +z direction?

Part B

What is the direction of the force exerted on this particle by a magnetic field with magnitude 2.00T in the +z direction?

Please, enter your answer as a counterclockwise angle from the +y direction to the direction of the force.

Part C

What is the magnitude of the force exerted on this particle by a magnetic field with magnitude 2.00T in the +z direction?

Explanation / Answer

We take the unit vectors as i for x-axis, j for y-axis and k for z-axis.

Charge on particle, q = -2.50 x 10-8 C

Magnitude of the velocity of the particle = 40 x 103 m/s

Angle of the velocity with the x-axis = 50o

So, in vector notation, the velocity of the particle is

v = 40000 * cos50 (i) + 40000 * sin50 (j)

v = 4 x 104 * (0.643(i) + 0.766(j))

Part A

The magnetic field has a magnitude of 2.00 T and acts in the +z direction. So the field vector is

B = 2(k)

The force on the particle is given by multiplying the charge to the cross product of velocity and magnetic field.

F = q (v X B)

F = -2.5 x 10-8 * (4 x 104 * (0.643(i) + 0.766(j)) X 2(k))

F = -20 x 10-4 * (0.643(-j) + 0.766(i))

F = -2 x 10-3 * (0.766(i) + 0.643(-j))

which can also be written as

F = -2 x 10-3 * (sin50(i) + cos50(-j))

Hence, the magnitude of the force can be given as

|F| = 2 x 10-3 * (sin250 + cos250)1/2 = 2 x 10-3 N

Ans. 2 x 10-3 N

Part B

We see that the unit vector in which this force acts is given by (sin50(-i) + cos50(j))

This means that the unit vector lies in the 2nd quadrant and makes an angle of 50o with the positive y-axis.

Ans. 50o CCW with +y direction.

Part C

Same as Part A if the field acts has a magnitude of 2 T in the +z direction.

If the field acts in the -z direction on the other hand, the magnitude of the force remains the same again as the magnitude of the field is not changing, but the direction of the force changes and becomes opposite to the direction in which it acts in Part A and B.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.