Question

The electrons in the beam of a television tube have a kinetic energy of 2.70 10-15 J. Initially, the electrons move horizontally from west to east. The vertical component of the earth's magnetic field points down, toward the surface of the earth, and has a magnitude of 1.00 10-5 T.
A) In what direction are the electrons deflected by this field component?
B) What is the magnitude of the acceleration of an electron in part (a)?

Answer 1

Answer:

[a]. South direction

(b). 1.354 × 10^14 m/s^2

Explanation:

In order to to be able to solve this question effectively, one must be familiar with physics concepts such as kinetic energy[1/2 ×mass × velocity,v²], magnetic force[ magnetic force = charge × velocity × magnetic field × sin Ф] and Newton's law of motion especially the second law of Newton[ Force = mass × acceleration].  

(a). The direction are the electrons deflected by this field component = q( v × B) = - (  i × -k) = -( - ( i × k) = - (- (- j) = -j. The -j is in the south direction.

Note that  i × k is being replaced by -j.

(b). In order to determine the magnitude of the acceleration of an electron in part (a), the first thing to do is to calculate the velocity from the equation for kinetic energy.

Thus, velocity = √ [(2 × 2.7 × 10⁻¹⁵) ÷ 9.1 × 10⁻³¹] = 7.7 × 10^7 m/s.

The next thing is to determine the net force.

Therefore, the net force = 1.6 × 10^-19 × 7.7 × 10^7 m/s × 1 × 10^-15 sin 90°.

= 1.232 × 10^ -16N.

Hence, Net Force = mass × acceleration.

1.232 × 10^ -16N = 9.1× 10^-31 × acceleration.

The acceleration = 1.232 × 10^ -16N/ 9.1× 10^-31 = 1.354 × 10^14 m/s^2.