A uniform magnetic field, find the magnitude of the force exerted by the field on a proton?

Question

Part 1:

A uniform magnetic field of magnitude 1.55 T acts in the positive z direction. Find the magnitude of the force exerted by

the field on a proton if the proton's velocity is (1.9 Mm/s) ^j. Answer in units of pN.

Part 2:

Find the magnitude of the force if the proton's velocity is (4Mm/s) ^j. Answer in units of pN.

Part 3:

Find the magnitude of the force for a velocity

of (6.4 Mm/s) ^k. Answer in units of pN.

Part 4:

Find the force if the velocity is (2 Mm/s) ^i +

(2 Mm=s) ^j. Answer in units of pN.

Answer 1

The i and j directions are both perpendicular to the z direction. If the proton travels in the z direction, the force is zero. The force exerted on the proton is given by F = qvB sin θ. Sine of 90 is 1. Sine of 0, 180, or 360 is 0.

Part 1:

F = qvB = (1.60 x 10^-19 C) x (1.90 x 10^6 m/s) x 1.55 T

= 0.471 pN

Part 2:

F = qvB = (1.60 x 10^-19 C) x (4.00 x 10^6 m/s) x 1.55 T

= 0.992 pN

Part 3:

F = Zero.

Part 4:

F = qvB = (1.60 x 10^-19 C) x (2.00 x 10^6 m/s) x 1.55 T

= 0.496 pN

Answer 2

a moving charge in a magnetic field has a force exerted on Of given by

F = q*(V x B) where x signifies the cross product.

In parts 1 and 2 the V is perpendicular to B so the force is simply

F = q*V*B you can substitute the correct values

In part 3,  V is parallel to B so the cross product is zero and

F = 0

In part 4 you need to use the cross product formula and note that

j x k = i

i x k = -j

Again, you supply the arithmetic.