Physics, asked by sjkskilled1037, 2 months ago

An electron moves north at a velocity of 4.5 × 104 m/s and has a magnetic force of 7.2 × 10-18 n exerted on it. if the magnetic field points upward, what is the magnitude of the magnetic field? use 1.60 × 10–19 c for the magnitude of the charge of an electron

Answers

Answered by maheshwari7127
0

Answer:

An electron moves north at a velocity of 4.5 × 104 m/s and has a magnetic force of 7.2 × 10-18 n exerted on it. if the magnetic field points upward, what is the magnitude of the magnetic field? use 1.60 × 10–19 c for the magnitude of the charge of an electron

Answered by abhi178
2

Given info : An electron moves north at a velocity of 4.5 × 104 m/s and has a magnetic force of 7.2 × 10-18 n exerted on it. if the magnetic field points upward,

To find : the magnitude of the magnetic field is ...

solution : velocity of electron, v = 4.5 × 10⁴ j^ m/s

[ electron is directed along North so in vector form j indicates its direction]

magnetic force, F = 7.2 × 10^-18 N

magnetic field points upward.

let B is magnetic field then in vector form , B k^

we see angle between magnetic field and velocity of electron is 90°

so, θ = 90°

charge on electron, q = 1.6 × 10^-19 C

we know,

F = BqV sinθ

⇒7.2 × 10^-18 = B × 1.6 × 10^-19 × 4.5 × 10⁴ × sin90°

⇒B = (7.2 × 10^-18)/(1.6 × 10^-15 × 4.5 )

= 10¯³ T

Therefore the magnitude of magnetic field is 10¯³ T.

also read similar questions : An electron is moving vertically upwards with a speed of 2.0*10^8 ms^-1,magnetic field is 0.50 N A^-1 m^-1.Calculate the...

https://brainly.in/question/53471

A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of ...

https://brainly.in/question/5883822

Similar questions