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3 years ago

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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?

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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