## Direction of Induced Current in a Solenoid

Contents

The approach for looking at the direction of current in a solenoid is the same the approach described above. The only difference being that in a solenoid there are a number of loops of wire so the magnitude of the induced emf will be different. The flux would be calculated using the surface area of the solenoid multiplied by the number of loops.

**Remember:** the directions of currents and associated magnetic fields can all be found using only the Right Hand Rule. When the fingers of the right hand are pointed in the direction of the magnetic field, the thumb points in the direction of the current. When the thumb is pointed in the direction of the magnetic field, the fingers point in the direction of the current.

The direction of the current will be such as to oppose the change. We would use a setup as in this sketch to do the test:

In the case where a north pole is brought towards the solenoid the current will flow so that a north pole is established at the end of the solenoid closest to the approaching magnet to repel it (verify using the Right Hand Rule):

In the case where a north pole is moving away from the solenoid the current will flow so that a south pole is established at the end of the solenoid closest to the receding magnet to attract it:

In the case where a south pole is moving away from the solenoid the current will flow so that a north pole is established at the end of the solenoid closest to the receding magnet to attract it:

In the case where a south pole is brought towards the solenoid the current will flow so that a south pole is established at the end of the solenoid closest to the approaching magnet to repel it:

### Tip:

An easy way to create a magnetic field of changing intensity is to move a permanent magnet next to a wire or coil of wire. The magnetic field must increase or decrease in intensity *perpendicular* to the wire (so that the magnetic field lines “cut across” the conductor), or else no voltage will be induced.

### Tip:

The induced current generates a magnetic field. The induced magnetic field is in a direction that tends to cancel out the change in the magnetic field in the loop of wire. So, you can use the Right Hand Rule to find the direction of the induced current by remembering that the induced magnetic field is opposite in direction to the change in the magnetic field.

### Induction

Electromagnetic induction is put into practical use in the construction of electrical generators which use mechanical power to move a magnetic field past coils of wire to generate voltage. However, this is by no means the only practical use for this principle.

If we recall, the magnetic field produced by a current-carrying wire is always perpendicular to the wire, and that the flux intensity of this magnetic field varies with the amount of current which passes through it. We can therefore see that a wire is capable of inducing a voltage *along its own length* if the current is changing. This effect is called *self-induction*. Self-induction is when a changing magnetic field is produced by changes in current through a wire, inducing a voltage along the length of that same wire.

If the magnetic flux is enhanced by bending the wire into the shape of a coil, and/or wrapping that coil around a material of high permeability, this effect of self-induced voltage will be more intense. A device constructed to take advantage of this effect is called an *inductor*.

Remember that the induced current will create a magnetic field that opposes the change in the magnetic flux. This is known as Lenz’s law.

## Example: Faraday’s Law

### Question

Consider a flat square coil with 5 turns. The coil is \(\text{0.50}\) \(\text{m}\) on each side and has a magnetic field of \(\text{0.5}\) \(\text{T}\) passing through it. The plane of the coil is perpendicular to the magnetic field: the field points out of the page. Use Faraday’s Law to calculate the induced emf, if the magnetic field is increases uniformly from \(\text{0.5}\) \(\text{T}\) to \(\text{1}\) \(\text{T}\) in \(\text{10}\) \(\text{s}\). Determine the direction of the induced current.

### Step 1: Identify what is required

We are required to useFaraday’s Law to calculate the induced emf.

### Step 2: Write Faraday’s Law

\[\mathcal{E}=-N\frac{\Delta\phi}{\Delta t}\] We know that the magnetic field is at right angles to the surface and so aligned with the normal. This means we do not need to worry about the angle that the field makes with the normal and \(\phi=BA\). The starting or initial magnetic field, \(B_i\), is given as is the final field magnitude, \(B_f\). We want to determine the magnitude of the emf so we can ignore the minus sign.

The area, \(A\), is the area of square coil.

### Step 3: Solve Problem

\begin{align*} \mathcal{E}&=N\frac{\Delta\phi}{\Delta t}\\ &=N\frac{\phi_f-\phi_i}{\Delta t} \\ &=N\frac{B_fA – B_iA}{\Delta t} \\ &=N\frac{A(B_f – B_i)}{\Delta t} \\ &=(5)\frac{(\text{0.50})^2(\text{1} – \text{0.50})}{\text{10}} \\ &=(5)\frac{(\text{0.50})^2(\text{1} – \text{0.50})}{\text{10}} \\ &=\text{0.0625}\text{ V} \end{align*}

The induced current is anti-clockwise as viewed from the direction of the increasing magnetic field.

## Example: Faraday’s Law

### Question

Consider a solenoid of 9 turns with unknown radius, \(r\). The solenoid is subjected to a magnetic field of \(\text{0.12}\) \(\text{T}\). The axis of the solenoid is parallel to the magnetic field. When the field is uniformly switched to \(\text{12}\) \(\text{T}\) over a period of 2 minutes an emf with a magnitude of \(-\text{0.3}\) \(\text{V}\) is induced. Determine the radius of the solenoid.

### Step 1: Identify what is required

We are required to determine the radius of the solenoid. We know that the relationship between the induced emf and the field is governed by Faraday’s law which includes the geometry of the solenoid. We can use this relationship to find the radius.

### Step 2: Write Faraday’s Law

\[\mathcal{E}=-N\frac{\Delta\phi}{\Delta t}\] We know that the magnetic field is at right angles to the surface and so aligned with the normal. This means we do not need to worry about the angle the field makes with the normal and \(\phi=BA\). The starting or initial magnetic field, \(B_i\), is given as is the final field magnitude, \(B_f\). We can drop the minus sign because we are working with the magnitude of the emf only.

The area, \(A\), is the surface area of the solenoid which is \(\pi r^2\).

### Step 3: Solve Problem

\begin{align*} \mathcal{E}&=N\frac{\Delta\phi}{\Delta t}\\ &=N\frac{\phi_f-\phi_i}{\Delta t} \\ &=N\frac{B_fA – B_iA}{\Delta t} \\ &=N\frac{A(B_f – B_i)}{\Delta t} \\ (\text{0.30})&=(9)\frac{(\pi r^2)(\text{12} – \text{0.12})}{\text{120}} \\ r^2 &= \frac{(\text{0.30})(\text{120})}{(9)\pi(\text{12} – \text{0.12})} \\ r^2&=\text{0.107175} \\ r&=\text{0.32}\text{ m} \end{align*}

The solenoid has a radius of \(\text{0.32}\) \(\text{m}\).

## Example: Faraday’s Law

### Question

Consider a circular coil of 4 turns with radius \(\text{3} \times \text{10}^{-\text{2}}\) \(\text{m}\). The solenoid is subjected to a varying magnetic field that changes uniformly from \(\text{0.4}\) \(\text{T}\) to \(\text{3.4}\) \(\text{T}\) in an interval of \(\text{27}\) \(\text{s}\). The axis of the solenoid makes an angle of \(\text{35}\)\(\text{°}\) to the magnetic field. Find the induced emf.

### Step 1: Identify what is required

We are required to useFaraday’s Law to calculate the induced emf.

### Step 2: Write Faraday’s Law

\[\mathcal{E}=-N\frac{\Delta\phi}{\Delta t}\] We know that the magnetic field is at an angle to the surface normal. This means we must account for the angle that the field makes with the normal and \(\phi=BA\cos(\theta)\). The starting or initial magnetic field, \(B_i\), is given as is the final field magnitude, \(B_f\). We want to determine the magnitude of the emf so we can ignore the minus sign.

The area, \(A\), will be \(\pi r^2\).

### Step 3: Solve Problem

\begin{align*} \mathcal{E}&=N\frac{\Delta\phi}{\Delta t}\\ &=N\frac{\phi_f-\phi_i}{\Delta t} \\ &=N\frac{B_fA\cos(\theta) – B_iA\cos(\theta)}{\Delta t} \\ &=N\frac{A\cos(\theta)(B_f – B_i)}{\Delta t} \\ &=(4)\frac{(\pi(\text{0.03})^2\cos(\text{35}))(\text{3.4} – \text{0.4})}{\text{27}} \\ &=\text{1.03} \times \text{10}^{-\text{3}}\text{ V} \end{align*}

The induced current is anti-clockwise as viewed from the direction of the increasing magnetic field.

### Real-life applications

The following devices use Faraday’s Law in their operation.

induction stoves

tape players

metal detectors

transformers

## Optional Project: Real-life applications of Faraday’s Law

Choose one of the following devices and do some research on the internet, or in a library, how your device works. You will need to refer to Faraday’s Law in your explanation.

induction stoves

tape players

metal detectors

transformers