We all have seen magnets and are very fond of playing with them. Also, we have seen how the same side of the magnets repels each other. But, the magnetic field is the field of current that attracts and repels the magnet. Furthermore, in this topic, you will learn about the magnetic field, magnetic field formula, its derivation, and solved examples.

**Magnetic Field**

When an electric current passes through a wire, it creates a magnetic field around it. Also, this magnetic field forms concentric circles around the wire. Furthermore, the direction of the magnetic field depends upon the direction of the current.

Moreover, we can determine it by using the ‘right-hand rule’, by pointing the thumb of your right hand in the direction of the current. Besides, the direction of magnetic field lines in the direction of your curled fingers. And the magnitude of filed depends on the amount of current, and the distance from the charge-carrying wire.

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**Magnetic Field Formula**

The magnetic field formula contains the \(constant^{\mu_{0}}\). This is known as permeability of free space and has a \(value^{\mu}_{0}\) = \(4\pi \times 10^{-7} (T \cdot m\)/ A). Besides, the unit of a magnetic field is Tesla (T).

Magnetic field magnitude = \(\frac{(permeability of free space) (current magnitude)}{2\pi (distance)}\)

B = \(\frac{\mu_{0}}{2\pi r}\)

**Derivation of the Formula**

B = refers to the magnetic field magnitude in Tesla (T)

\(\mu_{0}\) = refers to the permeability of free space (\(4\pi \times 10^{-7} T \cdot m/A\))

I = refers to the magnitude of the electric current in amperes (A)

r = refers to the distance in meters (m)

**Solved Examples on Magnetic Field**

**Example 1**

Find the magnitude of the magnetic field that is 0.10 m away from a wire carrying a 3.00 A current? Also, the current has a vector direction out of the page (or screen), then what is the direction of the magnetic field?

**Solution:**

By using the formula we calculate the magnetic field:

B = \(\frac{\mu_{0}}{2\pi r}\)

B = \(\frac{(4\pi \times 10^{-7} T \cdot m/A) (3.00 A)}{2\pi (0.10 m)}\)

B = \(\frac{(4\pi \times 10^{-7} T \cdot m/A) (3.00 A)}{2\pi (0.10 m)}\)

B = \(\frac{(4\pi \times 10^{-7}) (3.00)}{2\pi (0.10)}\)

B = \(\frac{(4\pi \times 10^{-7}) (3.00)}{2\pi (0.10)}\)T

B = \(\frac{4\pi (3.00)}{2\pi (0.10)} \times 10^{-7}\) T

B = \(2 (30.0) \times 10^{-7}\) T

B = 60.0 \(\times 10^{-7}\) T

B = 60.0 \(\times 10^{1-7}\) T

B = 60.0 \(\times 10^{-6}\) T

B = 6.00 \(\mu T\)

So, the magnitude of the magnetic field is 6.00 \(\times 10^{-6}\)T. Moreover, it can also be written as \(^{6.00\mu T}\) (micro-Tesla).

Besides, the direction of the magnetic field can be determined using the ‘right-hand rule’ by pointing the thumb of your right hand in the direction of the current. Also, the direction of the magnetic field lines in the direction of your curled fingers. Moreover, the current has a vector direction out of the page, and so your fingers will curl in the counter-clockwise direction. Hence, the magnetic field lines point in the counter-clockwise direction, forming circles around the wire.

**Example 2**

Assume that the magnitude of a magnetic field is 2.00 m away from a wire is 10.0 nT (nano-Tesla), so, what will be the electric current carried by the wire?

**Solution:**

Firstly, rearrange the magnetic field formula to find the magnitude of the electric current

B = \(\frac{\mu_{0}}{2\pi r}\) \(\rightarrow\) \(2\pi rB = \mu_{0}I\)

\(2\pi rB = \mu_{0}I\) \(\rightarrow\) I = \(\frac{2 \pi rB}{\mu_{0}}\)

I = \(\frac{2 \pi rB}{\mu_{0}}\)

Furthermore, the magnitude of the magnetic field is given in nano-Tesla. Also, the prefix nano means \(10^{-9}\), and 1 nT = \(10^{-9}\) T. So, the magnitude of the filed at the distance specified is thus:

B = 10.0 nT

B = (10.0 nT) \(\left ( \frac{10^{-9}T}{1 nT} \right )\)

B = (10.0 nT) \(\left ( \frac{10^{-9}T}{1 nT} \right )\)

B = 10.0 \(\times 10^{-9} T\)

B = 1.00 \(\times 10^{-8} T\)

Now the magnitude of the wire is

I = \(\frac{2 \pi rB}{\mu_{0}}\)

I = \(\frac{2\pi (2.00 m) (1.00 \times 10^{-8}T)}{(4\pi \times 10^{-7} T \cdot m/A) }\)

I = \(\frac{2 \pi}{4 \pi}\frac{(2.00 m) (1.00 \times 10^{-8}T)}{(4\pi \times 10^{-7} T \cdot m/A) }\)

I = \(\frac{2 \pi}{4 \pi} (2.00 \times 10^{-8+7})A\)

I = \(\frac{1}{2} (2.00 \times 10^{-}1)A\)

I = \(1.00 \times 10^{-1}A\)

I = 0.100 A

So, the magnitude of the electric current in the wire is 0.100 A.