life saver for the correct answer ..

Part A:

The largest possible acceleration occurs when the magnetic field is perpendicular to the direction the electron is moving.

Using the equation

F=qvB (for v is perpendicular to B)

and F=ma

We can solve for acceleration

a = qvB/m

 

Given that

q = 1.6 * 10^-19 C

m = 9.11*10^-31 kg

v=3*10^6 m/s

B= 8 * 10^-2 T

 

a = (1.6 * 10^-19)(3*10^6)(8 * 10^-2) / (9.11*10^-31) = 4.22*10^16 m/s^2

(Sorry if the math is wrong, I don't have a good calculator on hand)

 

Part B

The smallest acceleration occurs when the magnetic field is parallel to the velocity, in which case no force is exerted on the electron by the field. Thus the minimum acceleration is 0 m/s

 

Part C

Divide the acceleration from part A by 4, and we get 1.05 * 10^16 m/s^2

Now we backtrack using F=qvB and F=ma

Since time, the equation is altered to F=qvBsinθ , because the velocity is no longer perpendicular to the field.

 

Solving for θ

sinθ = ma/qvB

sinθ = 0.249

θ= sin-1 (0.249)

Once again, I don't have the calculator to solve this, but that should give you the answer.