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## An electron in a vacuum is first accelerated by a voltage of 81700 V and then enters a region in which there is a uniform magnetic field of

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An electron in a vacuum is first accelerated by a voltage of 81700 V and then enters a region in which there is a uniform magnetic field of 0.508 T at right angles to the direction of the electron’s motion. The mass of the electron is 9.11 × 10−31 kg and its charge is 1.60218 × 10−19 C. What is the magnitude of the force on the electron due to the magnetic field? Answer in units of N.

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Physics
3 years
2021-07-20T03:41:45+00:00
2021-07-20T03:41:45+00:00 2 Answers
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## Answers ( )

Answer:Explanation:After acceleration under potential difference , velocity v acquired can be calculated by the following expression

V e = 1/2 m v² ;

V is potential under which electron with mass m and charge e is accelerated to velocity v .

81700 x 1.60218 x 10⁻¹⁹ = .5 x 9.11 x 10⁻³¹ x v²

v² = 28737 x 10¹²

v = 169.52 x 10⁶ m /s

Force = Bev , B is magnetic field , e is charge on lectron and v is its velocity

= .508 x 1.60218 x10⁻¹⁹ x 169.52 x 10⁶

= 128 x 10⁻¹³ N.

Answer:Magnetic force is equal to

Explanation:We have given electron is accelerated with a potential difference of 81700 volt.

Magnetic field B = 0.508 T

Angle between magnetic field and velocity

Mass of electron

Charge on electron

By energy conservation.

Magnetic force on electron