Question

A proton moving in the positive x direction with a speed of 9.9 105 m/s experiences zero magnetic force. When it moves in the positive y direction it experiences a force of 1.6 10-13 N that points in the positive z direction. Determine the magnitude and direction of the magnetic field.

Answer 1

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

Step-by-step explanation:

In order to calculate the magnitude and direction of the magnetic field, you take into account the following equation for the magnetic force on the proton:

`\vec{F_B}=q\vec{v}\ X\ \vec{B}` (1)

v: speed of the proton = 9.9*10^5 m/s

q: charge of the proton = 1.6*10^-19C

B: magnetic field = ?

FB: magnetic force on the proton = 1.6*10^-13N

When the proton travels in the positive y direction (^j), you have that the proton experiences a force in the positive z direction (+^k). To obtain this direction of the magnetic force on the proton, it is necessary that the magnetic field points in the negative x direction, in fact, you have:

^j X (-^i) = -(-^k)=^k

To obtain the magnitude of the magnetic field you use:

`F_B=qvBsin90\°=qvB\\\\B=(F_B)/(qv)=(1.6*10^(-13)N)/((1.6*10^(-19)C)(9.9*10^5m/s))\\\\B=1.01T`

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction