An electron moving through a uniform magnetic field with a velocity of 2.0x10^6 m/s [up] experiences a maximum magnetic force of 5.1x10^-14 N [left]. CAlculate the magnitude and direction of the magnetic field.
To solve this problem, we can use the formula for magnetic force on a moving charged particle:
F = B * |q| * v * sin(θ),
where F is the magnetic force, B is the magnetic field, |q| is the magnitude of the charge, v is the velocity, and θ is the angle between the velocity and magnetic field.
From the information given in the problem, we know:
- The magnetic force, which is 5.1x10^-14 N [left].
- The velocity of the electron, which is 2.0x10^6 m/s [up].
To find the magnitude and direction of the magnetic field, we need to determine the angle θ. Since the magnetic force is directed to the left and the velocity is directed upwards, we can conclude that the angle θ between them is 90 degrees.
Since sin(90°) = 1, we can simplify the equation:
F = B * |q| * v
Now we can rearrange the equation to solve for B:
B = F / (|q| * v)
Let's substitute the given values into the equation and calculate the magnitude and direction of the magnetic field.
B = (5.1x10^-14 N) / (1.6x10^-19 C * 2.0x10^6 m/s)
B = 1.59 T (rounded to two decimal places)
Therefore, the magnitude of the magnetic field is 1.59 Tesla. To determine the direction, we can use the Right-Hand Rule. In this case, the magnetic field would be directed into the page.