you are given two circuits with two batteries of emf e and internal resistance r1 each. circuit a has the batteries connected in
1 answer:
Answer:
In which direction does the current in circuit A flow?
counterclockwise
<h2>What is the power dissipated by the resistor of resistance R2 for circuit A, given that E=10 V, R1=300ohms, and R2=5000ohms? </h2><h2>Calculate the power to two significant figures.</h2><h2>0.064W</h2><h2 /><h2>For what ratio of R1 and R2 would power dissipated by the resistor of resistance R2 be the same for circuit A and circuit B?</h2><h2>R1/R2 = 1 </h2><h2 /><h2>Under which of the following conditions would power dissipated by the resistance R2 in circuit A be bigger than that of circuit B? </h2><h2>Some answer choices overlap; choose the most restrictive answer.</h2><h2>R2>R1</h2><h2> </h2>Explanation:
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The magnitude of the magnetic moment due to the electron's motion is .
The magnetic pull and direction of a magnet or other object that produces a magnetic field are referred to as the magnetic moment in electromagnetism. Things that have magnetic moments include electromagnets, permanent magnets, various compounds, elementary particles like electrons, and a number of celestial objects (such as many planets, some moons, stars, etc).
The term "magnetic moment" really refers to the magnetic dipole moment of a system, which is the portion of the magnetic moment that can be represented by an equivalent magnetic dipole or a pair of magnetic north and south poles that are only very slightly apart. The magnetic dipole component is adequate for sufficiently small magnets or over sufficiently large distances.
Calculations:
radius=
velocity=
Working formula, M=N/A
=
=
M=
=
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Answer:
3.28 cm
Explanation:
To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:
r =
Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.
The mass and charge of a proton are:
m = 1.67 * 10^-27 kg
q = 1.6 * 10^-19 C
So, we get that the radius r will be:
r = = 0.0328 m, or 3.28 cm.
