An electron experiences a magnetic force of magnitude 4.60 x 10^-15 n when moving at an angle of 60.0° with respect to a magnetic field of magnitude 3.50 x 10^-3 t. find the speed of the electron.

9.49 × 10⁶ m/s

Explanation:

Data provided in the question:

Magnitude of Magnetic force, F = 4.60 × 10⁻¹⁵ N

Angle, θ = 60°

Magnitude of magnetic field, B = 3.50 × 10⁻³ T

Now,

we know

F = qVBsin(θ)

here ,

q is the charge of electron = 1.6 × 10⁻¹⁹ V

V is the speed of electron

Therefore,

4.60 × 10⁻¹⁵ = ( 1.6 × 10⁻¹⁹ ) × V × ( 3.50 × 10⁻³ ) × sin(60°)

or

V = 9.49 × 10⁶ m/s

if you answer my question i will answer yours

v=ir = 3x15 =45 v