Physics Question Pertaining To Electromagnetic Induction .

Physics: Electromagnetic Induction?

Yes, You can. In order to make a generator, i.e. to induce a current in a coil, all You need to do is make a device in which the lines of flux of a magnetic field go through the surface of the coil (surface around which the wire wrappings are made) and, while rotating, the flux has to change. In other words, the flux through the solenoid has to change as the permanent magnets (which create the flux) change position relative to the coil, as they rotate.

So, if You mount permanent magnets in such a manner that they create a certain flux through the coil, and then that flux changes as the permanent magnets rotate together with the shaft, the changing flux through the coil is what induces the current in it. Then You can connect an external electric circuit to the coil and use the electricity You created.

A typical example is a simple permanent magnet electric motor. It can behave as a generator, i.e. if You connect a voltmeter to its leads, and turn the rotor with Your fingers, the voltmeter will show some voltage.

Electromagnetic Induction. (physics help!!!)?

All of electricity production requires it - that's how you convert mechanical energy in to electric energy - by moving magnets you induce electric motion in the surrounding coils and you move the magnets by steam usually. Charging electric toothbrushes use it too (and plenty of things in between these).

TV and Radio will use electromagnetic induction in the sense that a changing magnetic field induces an electric field and an electric field induces a magnetic one. When these two sustain each other you get radiation - which the TV or radio pick up as signals.

Magnetic braking uses the principle too (the magnets induce eddy currents in the wheel which act to slow it down).

Physics question about electromagnetic induction!! Please please help!?

Answer … 1.42 mT …
According to Faraday’s law of induction, the EMF ℰ induced in a coil consisting
of N turns is … ℰ = - N ( dϕ / dt ) … where the flux of the magnetic is given by …
… ϕ = B A … where A = area of the flat circular coil … the rate of change in the
magnetic flux is therefore … dϕ / dt = A ( dB / dt ) because the area of the coil
is constant … it follows that … ℰ = - N A ( dB / dt ) = - N ( π r ² ) ( dB / dt ) …
using Ohm’s law ℰ = I R , we get … - N ( π r ² ) ( dB / dt ) = I R … so that the
induced current is … I = - ( π N r ² / R ) ( dB / dt ) … substituting the values, we
get … I = - [ π ( 105 ) ( 4.0 × 10 ⁻² m ) ² / ( 0.480 Ω ) ] ( 0.783 T / s ) = - 0.861 A
… where the negative sign indicates that the magnetic field produced by the
induced current opposes the change in flux produced by the primary field …
that is …with the coil seen from the top with the primary field going up, the
magnetic field produced by the induced current must be downward, so that the
induced current in the coil flows clockwise … using the formula B′ = μ₀ I / ( 2 r )
for the magnetic field at the center of one circular coil with radius r, we get for a
coil with N turns …
… B′ = N μ₀ I / ( 2 r )
…… = ( 105 ) ( 4 π × 10 ⁻⁷ T • m / A ) ( 0.861 A ) / [ 2 ( 4.0 × 10 ⁻² m ) ]
…… = 0.00142 T
…… = 1.42 mT

Physics electromagnetic induction problem need help with!!!?

By Faraday's Law the induced emf in a coil of wire:
V = -N(dΦ/dt), where N are the number of turns of wire and dΦ/dt id the rate of change of magnetic flux

By Ohm's Law:
V = IR

Definition of flux:
Φ = B ∙ A

Putting all of this together:
IR = -N(d(B ∙ A)/dt)

Since the area of the loop is constant and the magnetic field is perpendicular to the loop plane:
IR = -NA(dB/dt)

(0.09 A)(10 Ω) = -(67)(0.029 m)(0.124 m)(dB/dt)
dB/dt = -3.74 T/s

Physics problem.Electromagnetic induction?

It is the velocity component moving horizontally that causes the flux change from which the emf is generated..
V(h) = 1.0m/s x cos60 = 0.50m/s

The rate of change of flux-gathering area .. dA/dt = L x V(h) .. L= wire length, 0.05m
dA/dt = 0.05m x 0.50m/s = 0.025 m²/s

From Faraday's equation .. E = (-) N.d(B.A)/dt .. .. N=1 conductor cutting through field

E = B x dA/dt .. (B= constant) .. .. 0.50Wb/m² x 0.025m²/s .. .. .. ►E = 1.25^-2 V

Physics Electromagnetic Induction problem?

By the Farady's Law of Electromagnetic Induction,
Average induced e.m.f. = |-N (dΦ/dt)|
whereby N is the number of turns of the coil, dΦ/dt is the change in magnetic flux during the time interval with Φ=BA (B= magnetic field and A as the cross section area).

Therefore, in this case,
Average induced e.m.f. = |-N (dΦ/dt)|
= |-52 x [(0.5-0) x 4.6 10-3/ 0.46]|
= 0.26 V

I hope this will help. :)