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.

Answer 1

The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.

Step-by-step explanation:

Given that,

Speed
`v = 2.20*10^7\ m/s`

Acceleration
`a=1.90*10^(13)\ m/s^2`

We need to calculate the magnetic field

Using formula of magnetic field

`F=qvB`....(I)

Using newton's second law

`F= ma`....(II)

From equation (I) and (II)

`ma=qvB`

Put the value into the formula

`1.90*10^(13)*1.67*10^(-27)=1.6*10^(-19)*2.20*10^(7)*B`

`3.173*10^(-14)=1.6*10^(-19)*2.20*10^(7)*B`

`B=(3.173*10^(-14))/(1.6*10^(-19)*2.20*10^(7))`

`B=0.009014\ T`

We need to calculate the direction of the field

Using the right hand rule, point the right hand fingers along the velocity which is in the positive z direction.

Now, if we curl the fingers along the direction of magnetic field that is in the negative y direction, then the thumb will point in the positive x direction.

Hence, The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.