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Mekhanik [1.2K]
4 years ago
13

A) An electron is moving with a speed of 3.5 x105 m/s when it encounters a magnetic field of 0.60T. The direction of the magneti

c field makes an angle of60.0° with respect to the velocity of the electron. Whatis the magnitude of the magnetic force on the electron?B) Each second, 1.25 x 1019 electrons ina narrow beam pass through a small hole in a wall. The beam isperpendicular to the wall. Using Ampere’s law, determinethe magnitude of the magnetic field in the wall at a radius of0.750 m from the center of the beam.I know the answer to A is 2.9 x 10^-14 N and the answer to Bis 5.33 x 10^-7 T. Can anyone show me the steps? Ithink I've tried everything!! I will rate immediately!!!!
Physics
1 answer:
kirza4 [7]4 years ago
6 0

Given Information:  

v = 3.5x10⁵ m/s

θ = 60°

Magnetic field = 0.60 T

no. of electrons = 6.242x10¹⁸

radius = 0.750 m

Required Information:  

Magnetic force = F = ?

Magnetic field = B = ?

Answer:

Part A) F = 2.909x10⁻¹⁴ N

Part B) B = 5.34x10⁻⁷ T

Explanation:

Part A:

The magnetic force can be found using

F = qvBsin(θ)

Where q = 1.60x10⁻¹⁹ C and v is the speed of electron, B is the magnetic field and  θ is the angle between v and B.

F = 1.60x10⁻¹⁹*3.5x10⁵*0.60sin(60°)

F = 2.909x10⁻¹⁴ N

Part B:

From the Ampere's law, the magnetic field can be found using

B = μ₀I/2πr

Where μ₀ = 4πx10⁻⁷ is the permeability of free space, I is the current, and B is the magnitude of the magnetic field produced.

1 ampere of current has 6.242x10¹⁸ electrons per second so

I = 1.25x10¹⁹/6.242x10¹⁸

I = 2.0032 A

B = 4πx10⁻⁷*2.0032/2π*0.750

B = 5.34x10⁻⁷ T

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Using equilibrium of force in vertical direction

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k (0.331) = (2.82) (9.8)

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Case 2 : When both masses are hanged from spring

m = mass hanged from the spring = 3.09 + 2.82 = 5.91 kg

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3 years ago
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The work done is given by 742.5 J while the coefficient of kinetic friction between the block and the surface is 0.46.

<h3>What is the work done?</h3>

The work done is given by the use of the formula;

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If a pitch vibrates at 880 cycles, the octave below would vibrate at ____ cycles. 220 440 660 1760
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Answer:

If a pitch vibrates at 880 cycles, the octave below would vibrate at 440 cycles.

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Pitch of sound is defined as relative highest and the lowest point of sound. The pitch is dependent on frequency of vibration.

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