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
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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
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.
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