please check my answer
An electron moves north at a velocity of 9.8 * 10^4 m/s and has a magnetic force of 5.6 * 10^-18 N west exerted on it. If the magnetic field points upward, what is the magnitude of the magnetic field?
F=BqV
F= 5.6 * 10^-18N
q= 16 * 10^-19C
V= 9.8 * 10^4 m/s
B= ?
B= F/qV = 5.6 * 10^-18/ 1.6 * 10^-19 * 9.8 * 10^4
= 3.57 * 10^-4 T
is this correct?
correct
234
Well, I must admit, you've got the right formula! I'm impressed, but not as impressed as an electron moving at a velocity of 9.8 x 10^4 m/s! That's fast!
Now, let's see if your calculation is correct.
B = F/(qV)
Plugging in the given values:
B = (5.6 x 10^-18 N) / ((1.6 x 10^-19 C) x (9.8 x 10^4 m/s))
Doing the math, I get a result of approximately 3.57 x 10^-4 T.
Congratulations, your answer is correct! Give yourself a pat on the back!
Yes, your answer is correct. The magnitude of the magnetic field is 3.57 * 10^-4 T (Tesla). Well done!
Yes, your answer is correct. To determine the magnitude of the magnetic field, you can use the formula F = Bqv, where F is the magnetic force, q is the charge of the electron, v is the velocity, and B is the magnetic field. Rearranging the equation, you get B = F / (qv).
Substituting the given values, we have:
F = 5.6 * 10^-18 N (westward force)
q = 1.6 * 10^-19 C (charge of an electron)
v = 9.8 * 10^4 m/s (velocity of the electron)
Now you can calculate the magnetic field:
B = (5.6 * 10^-18 N) / ((1.6 * 10^-19 C) * (9.8 * 10^4 m/s))
B ≈ 3.57 * 10^-4 T (Tesla)
So, the magnitude of the magnetic field is approximately 3.57 * 10^-4 Tesla.