Answer:
Magnetic field can be used to produce current, infact a changing magnetic field can produce current.
A changing magnetic field in a loop causes the flux linked with the loop to change in turn generating a emf in the loop and therefore a current.
For a loop of area A and resistance R.
I =dPhi/dt/R
В. А
I = AcosФ/R .dB /dt
But it isn't reasonable to say that we can create a magnetic field by having a flow of current and this can be used to make more current because the current generated due to change in magnetic field created by increase/decrease in flow of current will be in a direction such that it will counter act the change in magnetic field caused by increase/decrease in current flow.(lenz's law).
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Ф= В. А
I = Acos dB Rd
C the correct but not sure?
Answer:
It is a measure of the electric force per unit charge on a test charge.
Explanation:
The magnitude of the electric field is defined as the force per charge on the test charge.
Since we define electric field as the force per charge, it will have the units of force divided by the unit of charge. This implies that the SI unit of electric field is given as Newton/Coulomb (N/C).
Answer:
¨Facts you didn´t know¨ or ¨unknown facts¨
Explanation:
The magnitude of the induced emf is given by:
ℰ = |Δφ/Δt|
ℰ = emf, Δφ = change in magnetic flux, Δt = elapsed time
The magnetic field is perpendicular to the loop, so the magnetic flux φ is given by:
φ = BA
B = magnetic field strength, A = loop area
The area of the loop A is given by:
A = πr²
r = loop radius
Make a substitution:
φ = B2πr²
Since the strength of the magnetic field is changing while the radius of the loop isn't changing, the change in magnetic flux Δφ is given by:
Δφ = ΔB2πr²
ΔB = change in magnetic field strength
Make another substitution:
ℰ = |ΔB2πr²/Δt|
Given values:
ΔB = 0.20T - 0.40T = -0.20T, r = 0.50m, Δt = 2.5s
Plug in and solve for ℰ:
ℰ = |(-0.20)(2π)(0.50)²/2.5|
ℰ = 0.13V
