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adoni [48]
3 years ago
5

A 420-turn circular coil with an area of 0.0650 m2 is mounted on a rotating frame, which turns at a rate of 22.3 rad/s in the pr

esence of a 0.0550-T uniform magnetic field that is perpendicular to the axis of rotation. What is the instantaneous emf in the coil at the moment that the normal to its plane is perpendicular to the field
Physics
1 answer:
Ivenika [448]3 years ago
7 0

Answer:

33.48 V

Explanation:

Parameters given:

Number of turns, N = 420

Magnetic field strength, B = 0.055 T

Area, A = 0.065 m²

Angular velocity, ω = 22.3 rad/s

EMF induced in a coil is given as:

EMF = -dΦ/dt

where Φ = magnetic flux

Magnetic flux, Φ, is given as:

Φ = B * N * A * cosωt

EMF = -d( B * N * A * cosωt) / dt

EMF = B * N * A * ω * sinωt

where ωt = 90°

Therefore:

EMF = 0.055 * 420 * 0.065 * 22.3 * sin90°

EMF = 33.48 V

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If the atomic mass of Sodium-18 is 18.02597 u, what is the binding energy?
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Answer:

<h3>The binding energy of sodium Na=<em>5.407791×10⁹J</em></h3>

Explanation:

<h3>Greetings !</h3>

Binding energy, amount of energy required to separate a particle from a system of particles or to disperse all the particles of the system. Binding energy is especially applicable to subatomic particles in atomic nuclei, to electrons bound to nuclei in atoms, and to atoms and ions bound together in crystals.

<h2>Formula : Eb=(Δm)c²</h2><h3>where:Eb= binding energy</h3><h3> .Δm= mass defect(kg)</h3><h3> c= speed of light 3.00×10⁸ms¯¹</h3><h2 /><h3><u>Given</u><u> </u><u>values</u></h3>
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<h3><u>required </u><u>value</u></h3>
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