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3 years ago

8

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 magnet

ic 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.

2 answers:

Vilka [71]3 years ago

8 0

Answer:

Magnetic force is equal to $1.37\times 10^{-11}N$

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 $m=9.11\times 10^{-31}kg$

Charge on electron $e=1.6\times 10^{-19}C$

By energy conservation.

$\frac{1}{2}mv^2=qV$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=1.6\times 10^{-19}\times 81700$

$v=169.4\times 10^6m/sec$

Magnetic force on electron

$F=qvBsin\Theta$

$F=1.6\times 10^{-19}\times 169.4\times 10^6\times 0.508\times sin90^{\circ}$

$=1.37\times 10^{-11}N$

Delicious77 [7]3 years ago

5 0

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.

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