Question

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.

Answer 1

Step-by-step 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 2

Magnetic force is equal to
`1.37* 10^(-11)N`

Step-by-step 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
`\Theta =90^(0)`

Mass of electron
`m=9.11* 10^(-31)kg`

Charge on electron
`e=1.6* 10^(-19)C`

By energy conservation.

`(1)/(2)mv^2=qV`

`(1)/(2)* 9.11* 10^(-31)* v^2=1.6* 10^(-19)* 81700`

`v=169.4* 10^6m/sec`

Magnetic force on electron

`F=qvBsin\Theta`

`F=1.6* 10^(-19)* 169.4* 10^6* 0.508* sin90^(\circ)`

`=1.37* 10^(-11)N`