Question

A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.20 107 m/s and experiences an acceleration of 1.90 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.

magnitude_____T

direction : +x, -x, +y, -y, +z, -z   

Answer 1

To find the magnitude

The magnitude of the magnetic field is going to be caluclated from the following formula for the force on a moving charge

where is the charge in Coulombs, is the velocity in metre per second, and is the strength of the magnetic field in Tesla.

Since we were given the acceleration, we have to multiply by the mass of the proton to get the applied force.

Rearranging this equation to get an expression for the magnetic field, we get

Substituting the numeric values, we get

The magnitude of the field is .

To find the direction

I have done this part separately to make the problem easier.

The first equation, written in vector form is

Since the acceleration, and hence, the force, is in the X-direction, the vector cross product of the velocity and the magnetic field should have this direction.

From the cyclic rule of vector cross-products, we know that

Therefore, the magnetic field is in the or negative Y-direction.

We can also use the right hand rule to find the direction of the magnetic field. To do this align the thumb of your right hand with the velocity, and the palm should point in the direction of the force. Then, whichever way your fingers point is the direction of the magnetic field.

Whichever way you orient your axes, you will find your fingers point in the negative Y-direction.