An electron moves at a speed of 5.4 106 m/s perpendicular to a constant magnetic field. The path is a circle of radius 1.0 10-3 m.
(a) What is the magnitude of the field?
T
(b) Find the magnitude of the electron's acceleration.
m/s2
force= Me*v^2/r= qBv
solve for r
To find the magnitude of the magnetic field, we can use the formula for the magnetic force on a charged particle moving in a magnetic field:
F = qvB
where F is the magnetic force acting on the electron, q is the charge of the electron (1.6 x 10^-19 C), v is the speed of the electron (5.4 x 10^6 m/s), and B is the magnetic field.
We know that the magnetic force provides the centripetal force required for the electron to move in a circle, which can be expressed as:
Fc = m * a
where Fc is the centripetal force, m is the mass of the electron (9.1 x 10^-31 kg), and a is the acceleration of the electron.
Since the centripetal force is provided by the magnetic force, we can equate these two forces:
qvB = m * a
Rearranging the equation to solve for the magnetic field B:
B = (m * a) / (q * v)
(a) To find the magnitude of the magnetic field, plug in the given values:
m = 9.1 x 10^-31 kg (mass of the electron)
q = 1.6 x 10^-19 C (charge of the electron)
v = 5.4 x 10^6 m/s (speed of the electron)
B = (9.1 x 10^-31 kg * a) / (1.6 x 10^-19 C * 5.4 x 10^6 m/s)
We can now solve for B. However, we need the value of the acceleration (a) to proceed.
(b) To find the magnitude of the electron's acceleration, we can use the centripetal acceleration formula:
a = v^2 / r
where v is the speed of the electron and r is the radius of the circular path.
Again, plug in the given values:
v = 5.4 x 10^6 m/s (speed of the electron)
r = 1.0 x 10^-3 m (radius of the circular path)
a = (5.4 x 10^6 m/s)^2 / (1.0 x 10^-3 m)
Now that we have the acceleration value, we can substitute it back into the equation for B and solve for the magnitude of the magnetic field.