Question

At one instant, v(vector) = 2.00 m/s(i), 4.00 m/s (j), 6.00 m/s (k) is the velocity of a proton in a uniform magnetic field : B(vector) = 2.00 T (i) 4.00 T (j)+ 8.00 T (k). Use the dot product to find the angle between vectors v and B , then use this angle to find the magnitude of the magnetic force using |FB| = |q| |~v| |B~ ||sin(theta)|.

Answer 1

To find the angle between vectors v and B, use the dot product formula. Then, use this angle to find the magnitude of the magnetic force using the given equation.

Step-by-step explanation:

To find the angle between the vectors v and B, we can use the dot product formula:

v · B = |v| |B| cos(θ)

Substituting the given values, we get:

2.00 * 2.00 + 4.00 * 4.00 + 6.00 * 8.00 = |v| * |B| * cos(θ)

Solving for cos(θ), we get:

cos(θ) = (2.00 * 2.00 + 4.00 * 4.00 + 6.00 * 8.00) / (|v| * |B|)

cos(θ) = 0.7778

Now, we can use this angle to find the magnitude of the magnetic force using the formula:

|FB| = |q| * |v| * |B| * sin(θ)

Substituting the given values, we get:

|FB| = (1.6 * 10^-19 C) * sqrt(2.00^2 + 4.00^2 + 6.00^2) * sqrt(2.00^2 + 4.00^2 + 8.00^2) * sin(acos(0.7778))

|FB| = 9.6 * 10^-13 N