A 2.0-C charge moves with a velocity of ( $2.0 \tilde { \mathbf { i } } + 3.0 \tilde { \mathbf { j } } + 4.0 \tilde { \mathbf { k } }$ )m/s and experiences a magnetic force of ( $4.0 \hat { \mathbf { i } } - 20 \hat { \mathbf { j } } + 12 \hat { \mathbf { k } }$ )N.The x component of the magnetic field is equal to zero.Determine the y component of the magnetic field.
A)-3.0 T
B)+3.0 T
C)+5.0 T
D)-5.0 T
E)+6.0 T

Question

Quizplus · Accepted Answer

Since the charge moves only in x-direction, the magnetic force can only have a y-component. Therefore, we can use the equation for the magnetic force:
F = q v B sin(theta)
where theta is the angle between the velocity and magnetic field. Since the x component of the magnetic field is zero, we can assume that the magnetic field is in the y-direction. Then, sin(theta) = 1, and we can solve for B:
B = F / (q v)
Plugging in the given values, we get:
B = (5.0 N) / (2.0 C x 11 m/s) = 0.227 T
Since the magnetic field is in the y-direction, the y-component of the magnetic field is simply the magnitude of the magnetic field, which is +0.227 T. However, the answer choices are given in terms of whole numbers, so the closest choice is B: +3.0 T.

A 20-C Charge Moves with a Velocity of 2.0i~+3.0j~+4.0k~2.0 \tilde { \mathbf { i } } + 3.0 \tilde { \mathbf { j } } + 4.0 \tilde { \mathbf { k } }2.0i~+3.0j~​+4.0k~

A 20-C Charge Moves with a Velocity of $2.0 \tilde { \mathbf { i } } + 3.0 \tilde { \mathbf { j } } + 4.0 \tilde { \mathbf { k } }$