In a particular location, the horizontal component of the earth\'s magnetic fiel

Question

In a particular location, the horizontal component of the earth's magnetic field has a magnitude of 1.80 times 10^-3 T. An electron is shot vertically straight up from the ground with a speed of 1.60 times 10^6 m/s. What is the magnitude of the acceleration caused by the magnetic force? Ignore the gravitational force acting on the electron. ____________ m/s^2 A proton with a speed of 3.57 times 10^6 m/s is shot into a region between two plates that are separated by a distance of 0.24 m. As the drawing shows, a magnetic field exists between the plates, and it is perpendicular to the velocity of the proton. What must be the magnitude of the magnetic field so the proton just misses colliding with the opposite plate? ___________ T

Explanation / Answer

Q1.
We know that, Force F = q*v*B

Given B = 1.80*10^-3 T (I cannot read it properly, but it appears to be 10^-3)

so a = F/m = q*v*B/m

=> a = (1.60*10^-19*1.60*10^6*1.80*10^-3)/(9.109*10^-31)

= 5.06*10^14 m/s^2

Pls post other ques separately.

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