Answer:
10.56×10^-13N
Explanation:
From the question, the electron and the magnetic field are perpendicular and make an of 90° with each other.
Force on the electron= qvBsinø
q= magnitude of charge on the electron= 1.6×10^-19C
v= velocity of the electron= 1.1×10^6 ms-1
B= magnetic field intensity= 6.0T
Ø= angle between electron and magnetic field=90°
F= 1.6×10^-19 × 1.1×10^6×6.0 × sin 90°
F= 10.56×10^-13N
Answer: 1.057×10^-12 N, towards the negative x axis
Explanation: An electron moving in a uniform magnetic field experiences a force.
This force is defined by the formulae below.
F =qvB sinx°
Where F = force on electronic charge
q = magnitude of electronic charge = 1.602×10^-19c
v = velocity of electron = 1.1×10^6 m/s
B = strength of magnetic field = 6.0T
x° = angle between the velocity of the electron and strength of magnetic field = 90° ( this is because the axis of the Cartesian plane are perpendicular)
By substituting the parameters, we have that
F = 1.602×10^-19 × 1.1×10^6 × 6 × sin 90
F = 1.602×10^-19 × 1.1×10^6 × 6 × 1
F = 1.057×10^-12 N
The direction of this force is gotten by using cross product on the direction of the velocity and the magnetic field.
Magnetic field is directed towards the negative z axis (-k) and velocity is directed towards the positive y axis (j)
j × (-k) = -i
Hence the force is directed towards the negative axis
