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Schach [20]
3 years ago
13

An electron inside of a television tube moves with a speed of 2.80×107 m/sm/s . It encounters a region with a uniform magnetic f

ield oriented perpendicular to its trajectory. The electron begins to move along a circular arc of radius 0.190 mm . What is the magnitude of the magnetic field?
Physics
1 answer:
kaheart [24]3 years ago
3 0

Answer:

The magnitude of the magnetic field is 0.83 T

Explanation:

Given :

Speed of electron inside television tube (<em>v</em>) = 2.8×10^7 m/s

Radius of electron moving path (r) = 0.190×10^-3m

According lorentz force in magnetic field is,

∴ F = q (v ×B)

Where, q = charge of electron, v = speed of electron, B = magnetic field and

F = force on electron.

when electron is moving in circular orbit the force is given by,

∴ F = mv^2/r

so we equate above both equation.

∴ mv^2/r = q(v × B)

∴ B = mv/qr

∴ B = 9.1×10^-31×2.8×10^7/1.6×10^-19×0.190×10^-3

∴ B = 0.83 Tesla

Therefore, the magnitude of the magnetic field is 0.83 T.

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F_{Nx}=F_x-f

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F_{Nx}=ma_x

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