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krek1111 [17]
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.
Physics
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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Answer:

The induced current is 26.7 mA

Explanation:

Given;

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The induced emf is calculated as;

emf = - \frac{d \phi}{dt} \\\\emf = -\frac{dB.A}{dt} \\\\emf = A (\frac{dB}{dt} )\\\\emf = 0.078(0.24)\\\\emf = 0.0187 \ V

The resistance of the loop = 0.7 Ω

The induced current is calculated as;

V = IR\\\\I = \frac{V}{R} = \frac{emf}{R} = \frac{0.0187}{0.7} = 0.0267 \ A = 26.7 \ mA

4 0
3 years ago
A parallel beam of light in air makes an angle of 43.5 ∘ with the surface of a glass plate having a refractive index of 1.68. Yo
aniked [119]

Answer:

a) 46.5º  b) 64.4º

Explanation:

To solve this problem we will use the laws of geometric optics

a) For this part we will use the law of reflection that states that the reflected and incident angle are equal

     θ = 43.5º

This angle measured from the surface is

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b) In this part the law of refraction must be used

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The index of air refraction is n₁ = 1

The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface

     θ_s = 90 - θ

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The angle with respect to the surface is

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Answer:

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Anastaziya [24]

Answer: 1.089\times 10^5\ A

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