We learned in lecture that the magnetic force on a charge q
moving with velocity v in a magnetic field ~B has the form
F = q~v × B
One important fact about the cross-product is that for two vectors A and B , the following is
true A × B = −B × A which may also be determined by using the right-hand-rule. A convenient
way to calculate a cross-product is to decompose the vectors involved into x,y and z components,
and then use the following basic cross-product results for the basis vectors,
ˆx × ˆy = ˆz , ˆy × ˆz = ˆx , ˆz × ˆx = ˆy
Using the result just stated, the cross-products in the opposite order produce a minus sign,
ˆy × ˆx = −ˆz , ˆz × ˆy = −ˆx , ˆx × ˆz = −ˆy
so one only needs to remember the first three identities above, and these other identities can be
easily remembered. For this problem, use these identities to calculate the magnetic force on a
particle with a charge q = 7μC in a magnetic field B = (5 × 10−5 T)ˆy. Do the calculation for the
following three different velocities:
(a) v = v0ˆx
(b) v = v0ˆy
(c) v = v0 (x+y/suare root 2) where v0 = 5m/s
To calculate the magnetic force on a particle with charge q moving with velocity v in a magnetic field B, we'll use the equation:
F = q * v x B
where "x" represents the cross product.
Given that the charge q = 7μC and the magnetic field B = (5 × 10^-5 T)ˆy, we need to calculate the magnetic force for three different velocities:
(a) v = v0ˆx
We're given v0 = 5 m/s. Substituting the values into the equation, we have:
F = (7μC) * (v0ˆx x B)
= (7μC) * (v0 * B * sinθ) * ˆz
Since v0 is in the x-direction (ˆx) and B is in the y-direction (ˆy), the angle θ between them is 90 degrees. The sin(90) = 1.
Thus, the magnetic force is:
F = (7μC) * (5 m/s * 5 × 10^-5 T * 1) * ˆz
= 0.175 × 10^-3 N * ˆz
(b) v = v0ˆy
Using the same process, but now v = v0ˆy, we have:
F = (7μC) * (v0ˆy x B)
= (7μC) * (v0 * B * sinθ) * -ˆx
Again, with v0 in the y-direction (ˆy) and B in the y-direction (ˆy), the angle θ between them is 0 degrees since they lie in the same direction. The sin(0) = 0.
Hence, the magnetic force is:
F = (7μC) * (5 m/s * 5 × 10^-5 T * 0) * -ˆx
= 0
The magnetic force in this case is zero.
(c) v = v0 (x+y/square root(2))
Using the same approach as before, but with v = v0 (x+y/square root(2)), we have:
F = (7μC) * (v0 (x+y/square root(2)) x B)
= (7μC) * (v0/square root(2)) * (B * sinθ) * (-ˆz + ˆx)
Again, B is in the y-direction (ˆy), so the angle θ between v and B is 45 degrees. The sin(45) = 1/square root(2).
Hence, the magnetic force is:
F = (7μC) * (5 m/s/square root(2)) * (5 × 10^-5 T * 1/square root(2)) * (-ˆz + ˆx)
= -0.0875 × 10^-3 N * ˆz + 0.0875 × 10^-3 N * ˆx
Therefore, the magnetic force on the particle with charge q in the given magnetic field for the three different velocities is:
(a) F = 0.175 × 10^-3 N * ˆz
(b) F = 0
(c) F = -0.0875 × 10^-3 N * ˆz + 0.0875 × 10^-3 N * ˆx