Question

An electron moves with a speed of 5.0 × 104 m/s perpendicular to a uniform magnetic field of magnitude 0.20 T. What is the magnitude of the magnetic force on the electron? (e = 1.60 × 10-19 C) Group of answer choices

Answer 1

1.6×10⁻¹⁵ N

Step-by-step explanation:

Using,

F = qvBsinФ........................ Equation 1

Where F = Force, q = charge of the electron, v = speed of the electron, B = Magnetic Field, Ф = angle between the magnetic Field and the electron speed.

Given: v = 5.0×10⁴, B = 0.2 T, q = 1.60×10⁻¹⁹ C, Ф = 90° ( Perpendicular)

F = 1.60×10⁻¹⁹ (5.0×10⁴)(0.2)sin90°

F = 1.6×10⁻¹⁵ N

Hence the magnitude of the magnetic force on the electron = 1.6×10⁻¹⁵ N

Answer 2

the magnitude of the magnetic force on the electron is 1.6 × 10⁻¹⁵N

Step-by-step explanation:

Let's recall magnetic force on moving charge as follows:

where:

F = magnetic force ( N )

B = magnetic field strength ( T )

q = charge of object ( C )

v = speed of object ( m/s )

θ = angle between velocity and direction of the magnetic field

The formula for the magnetic force is given by :

`F=qvB\ sin\theta`

Given that,

speed of proton = v = 5.0 × 10⁴ m/s

magnetic field strength = B = 0.20 T

charge of proton = q = 1.60 × 10⁻¹⁹ C

direction of speed = θ = 90°

`F=qvB, q is the charge on electronF=1.6* 10^(-19)* 5* 10^4* 0.2F=1.6* 10^(-15)\ N`

the magnitude of the magnetic force on the electron is 1.6 × 10⁻¹⁵N