The link shows an illustration of a battery and a resistor….the area & the magnetic field have no bearing on the current in the resistor (except in a sort of “existential” way, LOL!)…

To calculate the current through the resistor, one need only know the simple equation V=IR, by which the answer to your question is 9 = I x 20 ohms, which tells us the current is 0.45 amps and the power is 4.05 watts…

The above answer from “Amonynous” is incorrect – the area and magnetic field *do* have an impact on the current in the resistor, even if it is only slight. See Lenz’s Law, which states that a current is induced in a closed conducting loop if the magnetic flux through the loop is changing. In this case the magnetic flux is indeed changing through the loop.

First find the magnetic flux ( Φ ) through the formula Φ = ABcosθ

A is the area of the loop (in m^2)

B is the strength of the magnetic field (in T)

θ is the angle of the axis of the loop relative to the direction of the magnetic field.

The area of the loop is 20cm^2, which in SI units of meters is (20 x 10^-4)m^2, or 0.002m^2.

The final strength of the magnetic field is 0.65T.

The angle of the axis is parallel to the field (0 degrees), and cos of 0 is just 1.

So Φ = 0.002m^2 x 0.65 x 1, which equals 0.0013

Now use Faraday’s Law to calculate the induced EMF, which is ε = ΔΦ / Δt

The field changed from 0T to 0.65T, so since it started at 0T the value for ΔΦ 0.0013

The time elapsed ( Δt ) is 14ms, which in SI units is 0.014 seconds.

So ε = 0.0013 / 0.014, which equals 0.09286

ε is also equivalent to the change in voltage, so in this case we’re looking at 0.09286V.

Lenz’s Law states that the direction of the induced current is the direction that would create a magnetic field that *opposes* the magnetic flux. In this case the magnetic flux is increasing out of the page, and so the current will be in the direction that will oppose that – or into the page. Using the right hand rule you can see that to create a field which opposes this current, the induced current must be in the clockwise direction. This is the same direction as the current being provided by the battery, so simply add them together: 9.0V + 0.09286V = 9.09286V

Now use Ohm’s Law ( I = V / R ) to calculate the current flowing through the resistor. I = 9.09286V / 20ohm, which then equals 0.45464 amps.

If you reduce this to only two significant figures then you will come up with the same answer as just using Ohm’s Law from the beginning. However if they’re incorporating a magnetic field then they are asking a question about electromagnetic induction, in which case the additional 0.0464A is a result of the changing magnetic field.

The link shows an illustration of a battery and a resistor….the area & the magnetic field have no bearing on the current in the resistor (except in a sort of “existential” way, LOL!)…

To calculate the current through the resistor, one need only know the simple equation V=IR, by which the answer to your question is 9 = I x 20 ohms, which tells us the current is 0.45 amps and the power is 4.05 watts…

The above answer from “Amonynous” is incorrect – the area and magnetic field *do* have an impact on the current in the resistor, even if it is only slight. See Lenz’s Law, which states that a current is induced in a closed conducting loop if the magnetic flux through the loop is changing. In this case the magnetic flux is indeed changing through the loop.

First find the magnetic flux ( Φ ) through the formula Φ = ABcosθ

A is the area of the loop (in m^2)

B is the strength of the magnetic field (in T)

θ is the angle of the axis of the loop relative to the direction of the magnetic field.

The area of the loop is 20cm^2, which in SI units of meters is (20 x 10^-4)m^2, or 0.002m^2.

The final strength of the magnetic field is 0.65T.

The angle of the axis is parallel to the field (0 degrees), and cos of 0 is just 1.

So Φ = 0.002m^2 x 0.65 x 1, which equals 0.0013

Now use Faraday’s Law to calculate the induced EMF, which is ε = ΔΦ / Δt

The field changed from 0T to 0.65T, so since it started at 0T the value for ΔΦ 0.0013

The time elapsed ( Δt ) is 14ms, which in SI units is 0.014 seconds.

So ε = 0.0013 / 0.014, which equals 0.09286

ε is also equivalent to the change in voltage, so in this case we’re looking at 0.09286V.

Lenz’s Law states that the direction of the induced current is the direction that would create a magnetic field that *opposes* the magnetic flux. In this case the magnetic flux is increasing out of the page, and so the current will be in the direction that will oppose that – or into the page. Using the right hand rule you can see that to create a field which opposes this current, the induced current must be in the clockwise direction. This is the same direction as the current being provided by the battery, so simply add them together: 9.0V + 0.09286V = 9.09286V

Now use Ohm’s Law ( I = V / R ) to calculate the current flowing through the resistor. I = 9.09286V / 20ohm, which then equals 0.45464 amps.

If you reduce this to only two significant figures then you will come up with the same answer as just using Ohm’s Law from the beginning. However if they’re incorporating a magnetic field then they are asking a question about electromagnetic induction, in which case the additional 0.0464A is a result of the changing magnetic field.