An electron (m=9.11 * 10^-31) enters a uniform magnetic field B=1.8T and moves perpendicular to the field lines with a speed v=0.92c. what is the radius of curvature of its path?
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To find the radius of curvature of the electron's path, you can use the formula for the radius of curvature of a charged particle moving in a magnetic field:
r = (m * v) / (q * B)
where:
r is the radius of curvature
m is the mass of the electron (9.11 * 10^-31 kg)
v is the velocity of the electron (0.92c)
q is the charge of the electron (-1.6 * 10^-19 C)
B is the magnetic field strength (1.8 T)
First, let's calculate the values for the variables in the formula:
m * v = (9.11 * 10^-31 kg) * (0.92 * speed of light)
= (9.11 * 10^-31 kg) * (0.92 * 3 * 10^8 m/s)
= 24.84 * 10^-23 kg.m/s
Now, substitute the values into the formula:
r = (24.84 * 10^-23 kg.m/s) / ((-1.6 * 10^-19 C) * (1.8 T))
= (24.84 * 10^-23) / (1.44 * 10^-19)
= 17.25 * 10^-4 m
= 1.725 * 10^-3 m
Therefore, the radius of curvature of the electron's path is 1.725 * 10^-3 meters.