An electron in the beam of a cathod-ray tube is accelerated by a potential difference of 2.14 kV . Then it passes through a region of transverse magnetic field, where it moves in a circular arc with a radius of 0.170 m . What is the magnitude of the field?

true

explanation:

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary 

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft