Question

A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.50 107 m/s and experiences an acceleration of 2.20 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

9.175 x 10∧-3

Step-by-step explanation:

since acceleration is in positve X direction the magnetic field must be in negative Y direction

acceleration to right hand thumb rule.

B = fm/qvsinO = ma/qvsin0

B = (1.67 x 10∧-27)(2.20 x 10∧13) / (1.60 x 10∧-19)(2.50 x 10∧7)sin90

B = 3.67 x 10∧-14 / 4 x 10∧-12

= 9.175 x 10∧-3

B = 9.175 x 10∧-3 in negative Y direction

Answer 2

A) B = 0.009185 T

B) Drection is negative y-direction

Step-by-step explanation:

A) We are given;

Speed(v) = 2.5 x 10^(7) m/s

Acceleration (a) = 2.2 x 10^(13) m/s²

We also know that charge of proton(q) = 1.6 x 10^(-19)

Mass of proton(m) = 1.67 x 10^(-27)

Now, Since the proton is moving by circular motion, this force is equal to the centripetal force which is given as;

F = qvBsinθ = ma

Since perpendicular, θ = 90°

And so, sinθ = sin 90 = 1

Thus, qvB = ma

Making B the subject gives;

B = ma/qv

B = (1.67 X 10^(-27) X 2.2 X 10^13)) / (1.6 X 10^(-19) X 2.5 X 10^(7))

= 0.009185 T

B) By use of Flemings right hand rule, we can see that the middle finger points toward negative y-direction, so the magnetic field is in the negative y-direction