# Calculating Induced Current in a Coil of Wire

• roam
In summary: Remember to convert units when necessary.In summary, a flat coil of wire with 20 turns and an area of 50 cm2 is placed in a uniform magnetic field that increases from 2.0 T to 6.0 T in 2.0 s. The coil has a total resistance of 0.40 Ω. Using Faraday's law of induction, we can find the induced current by finding the change in magnetic flux over time. The correct equation to use is I = (20*0.5*4pi*10^-7*(6-2))/0.4 = 0.5 A. It is important to remember to use the correct units when solving problems.
roam

## Homework Statement

A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 2.0 T to 6.0 T in 2.0 s. If the coil has a total resistance of 0.40 Ω , what is the magnitude of the induced current?

## Homework Equations

$$B= \mu_0 \frac{N}{l} I$$

$$\Phi_B = \int B.dA$$

$$L=\mu_0 N^2 A$$

$$L=\frac{\Phi_B}{I}$$

## The Attempt at a Solution

Using the above equations and rearranging we get

$$I=\frac{\Phi_B}{\mu_0N^2A}$$

$$I=\frac{\Phi_B}{(4\pi \times 10^{-7}) 20^2 (0.5)}$$

Now how do I evaluate the flux? I use $$\Phi_B = LI$$, then the I's will cancel and I can't solve the problem...

Are there simpler ways of solving this problem? How do I need to make use of dB/dt which is given to us?

You don't need to use the inductance equations (i.e. the equations with 'L') here; you just need Faraday's law of induction:
$$\frac{d\Phi_B}{dt} = - \mathcal{E}$$

where the 'E' is the EMF.

Remember that currents are only induced by changing magnetic fields, just like magnetic fields are only induced by changing electric fields.

zhermes said:
You don't need to use the inductance equations (i.e. the equations with 'L') here; you just need Faraday's law of induction:
$$\frac{d\Phi_B}{dt} = - \mathcal{E}$$

where the 'E' is the EMF.

Remember that currents are only induced by changing magnetic fields, just like magnetic fields are only induced by changing electric fields.

This equation doesn't take into acoount the number of turns in the coil, but I've tried it

$$\frac{-4}{2} = -I (0.4)$$

This gives us I=5 A, but the correct answer must be 0.5A. I had all the units in the SI units so what's the problem?

roam said:
This equation doesn't take into acoount the number of turns in the coil
Remember that $$\Phi_B = \int_A B dA$$ is the total flux, not just the magnetic field.

The flux will be the area times the number of loops, times the magnetic field.

zhermes said:
Remember that $$\Phi_B = \int_A B dA$$ is the total flux, not just the magnetic field.

The flux will be the area times the number of loops, times the magnetic field.

so that flux is A.N.B, but we don't know what the magnetic field is, the only thing we are given is dB/dt.

$$\Phi_B=A \times N \time B= (0.5) (20) (\mu_0 \frac{N}{l} I)$$

this doesn't work because we don't know what I is (we're looking for I), do you see the problem?

roam said:
so that flux is A.N.B, but we don't know what the magnetic field is, the only thing we are given is dB/dt
You are given both the magnetic field, and the change in it. $$B_0 = 2T$$, $$B_1 = 6T$$, $$\Delta t = 2s$$

roam said:
$$\Phi_B=A \times N \time B= (0.5) (20) (\mu_0 \frac{N}{l} I)$$
Again, this is not the right equation.

$$\frac{d \Phi_B}{dt} = - \mathcal{E} = I R = \frac{d}{dt} A N B = A N \frac{dB}{dt}$$
You know everything except I.

zhermes said:
Again, this is not the right equation.

$$\frac{d \Phi_B}{dt} = - \mathcal{E} = I R = \frac{d}{dt} A N B = A N \frac{dB}{dt}$$
You know everything except I.

$$I R = \frac{d}{dt} A N B = A N \frac{dB}{dt}$$

$$I = \frac{2 \times 20 \times 0.5}{0.4} = 50$$

So if this is the correct equation, why I'm still not getting the right answer?

roam said:
$$I R = \frac{d}{dt} A N B = A N \frac{dB}{dt}$$

$$I = \frac{2 \times 20 \times 0.5}{0.4} = 50$$

So if this is the correct equation, why I'm still not getting the right answer?

50 square centimeters is not 0.5 square meters.

## 1. How do you calculate the induced current in a coil of wire?

The induced current in a coil of wire is calculated using the equation I = -N (ΔΦ / Δt), where I is the induced current, N is the number of turns in the coil, ΔΦ is the change in magnetic flux, and Δt is the change in time.

## 2. What is magnetic flux and how does it relate to induced current?

Magnetic flux is a measure of the total magnetic field passing through a given area. It is directly proportional to the induced current in a coil of wire, meaning that a change in magnetic flux will result in an induced current in the coil.

## 3. How does the number of turns in a coil affect the induced current?

The number of turns in a coil directly affects the induced current. As the number of turns increases, the induced current also increases. This is because a larger number of turns allows for a greater magnetic field to be generated, resulting in a stronger induced current.

## 4. Can the induced current in a coil be negative?

Yes, the induced current in a coil can be negative. This occurs when the change in magnetic flux is negative, meaning the magnetic field is decreasing. The negative sign in the equation represents the direction of the induced current, which will be opposite to the direction of the change in magnetic flux.

## 5. What factors can cause a change in magnetic flux and therefore, induce a current in a coil?

There are several factors that can cause a change in magnetic flux, including a change in the strength of the magnetic field, a change in the area of the coil, or a change in the orientation of the coil with respect to the magnetic field. Additionally, moving a magnet towards or away from the coil, or changing the current in a nearby wire can also cause a change in magnetic flux and induce a current in the coil.

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