An electron moves to the left with a velocity of 9.6x10^5 m/s and has a magnetic force of 2.6x10^-15 N exerted on it. What is the magnitude of the magnetic field? Assume the charge has a value of 1.60x10^-19 C
To find the magnitude of the magnetic field, we can use the formula for the magnetic force on a charged particle moving through a magnetic field:
F = qvB
Where:
- F is the magnetic force on the particle
- q is the charge of the particle
- v is the velocity of the particle
- B is the magnitude of the magnetic field
In this case, we are given:
- F = 2.6x10^-15 N
- q = 1.60x10^-19 C
- v = 9.6x10^5 m/s
Plugging these values into the formula, we get:
2.6x10^-15 N = (1.60x10^-19 C)(9.6x10^5 m/s)(B)
Now, let's solve for B by dividing both sides of the equation by (1.60x10^-19 C)(9.6x10^5 m/s):
B = 2.6x10^-15 N / (1.60x10^-19 C)(9.6x10^5 m/s)
B = 170.83 T
Therefore, the magnitude of the magnetic field is 170.83 Tesla (T).