Question

If you live in Melbourne, Australia, the local magnetic field has a strength of about 4x10-5 T. The magnetic field vector is directed northward, making an angle of 30 deg above the horizontal. An electron in Melbourne is moving parallel to the ground, in the west direction, at a speed of 9x105 m/s. What are the magnitude and direction of the magnetic force on the electron

Answer 1

`5.76* 10^(-18)\ \text{N}` perpendicular to the velocity and magnetic field

Step-by-step explanation:

B = Magnetic field =
`4* 10^(-5)\ \text{T}`

`\theta` = Angle the magnetic field makes with the horizontal =
`30^(\circ)`

v = Velocity of electron =
`9* 10^5\ \text{m/s}`

q = Charge of electron =
`1.6* 10^(-19)\ \text{C}`

Magnetic force is given by

`F=qvB\sin\theta\\\Rightarrow F=1.6* 10^(-19)* 9* 10^5* 4* 10^(-5)\sin30^(\circ)\\\Rightarrow F=2.88* 10^(-18)\ \text{N}`

The magnitude of the magnetic force is
`2.88* 10^(-18)\ \text{N}` and the direction is perpendicular to the velocity and magnetic field.