When a current is passed through a conductor, a magnetic field is produced. The same happens with a solenoid when an electrical current passes through it. When a current passes through a solenoid, then it becomes an electromagnet. The formula for the magnetic field in a solenoid is \(B = {\mu _0}nI.\)

Furthermore, do you know that solenoids are commonly used in electronics and electromagnets as a way to create a uniform magnetic field? Do you know how does the loop of the current-carrying solenoid produces a magnetic field? In magnetics, to calculate the magnetic field of a highly symmetric configuration carrying a steady current, we use Ampere’s Circuital Law. In this article, we will discuss magnetic field inside a solenoid, solenoid formula, magnetic field due to a current in a solenoid and magnetic field of solenoid formula.

What is a Solenoid?

A solenoid is a long wire that is wounded in the form of a helix, and with the help of this configuration, we can produce a reasonably uniform magnetic field. We can consider each turn of the solenoid as a circular loop when the turns are closely spaced, and the net magnetic field is the vector sum of the fields resulting from all the turns. The interior field becomes more uniform and as the length of the solenoid increases.

When the turns are closely spaced, and the length is much greater than the radius of the turns, then we can approximate it as the ideal solenoid. The external field is zero, and the interior field is uniform over a great volume in this case. Ampere’s law is used to obtain an expression for the magnetic field in an ideal solenoid.

Ampere’s Circuital Law

Let us consider the boundary, which is made up of the number of small line elements. Consider one such element of length \((d\vec l).\) Take the tangential component of the magnetic field produced at this element and multiply it by the length of that element \((d\vec l).\) All such products are added together and consider the limit as the lengths of elements get smaller, and their number gets larger. The sum then tends to an integral. Ampere’s law states that the line integral of the magnetic field around a closed loop is proportional to the amount of current that is enclosed by the loop.
\(\oint {\vec B} \cdot d\vec l = {\mu _0}{I_{{\text{enc }}}}\quad \ldots …\left({\text{i}} \right)\)
Where \({\mu _0}\) is the constant of proportionality, and it is the permeability of space and has a value of
\({\mu _0} = 4\pi \times {10^{ – 7}}\;\rm{T – m/A}\)

Unit: Magnetic field is a vector quantity and has a SI unit of tesla \(\left( \rm{T} \right).\)

Direction: We can get the direction of the magnetic field with the help of the right-hand thumb rule.

Magnetic Field Lines in a Solenoid

To understand the nature of magnetic field lines inside the solenoid, let us take two similar currents carrying circular loops kept co-axially near to each other. The magnetic field produced due to each coil is shown below figure. At any point \(P\) at some distance (more than the radius of the coil) from the axis, we can observe that the magnetic field due to each coil is in the opposite direction.

If the distance between both coils will be very small, they will almost cancel each other, and the resultant magnetic field will be very weak. Take another point \(Q\) at the axis or close to the axis. At this point, the magnetic field due to both the coil is in the same direction, so the resultant field will be stronger in nature.

When a stake of the circular coil is arranged closely, it forms a solenoid. At any outside point, the field due to any coil will be opposed by other neighbouring coils. This will cause a tendency to have zero magnetic fields outside the coil. On the other hand, at any inside point, the field due to each coil will be additive in nature. This cause tendency to have a stronger and uniform magnetic field inside the coil. The tendency to have zero fields outside and a uniform field inside become more and more effective when the solenoid is tightly wounded and have a large number of coils.

For an ideal solenoid, the magnetic field at any outside point will be zero, and inside it will be uniform.

Formula of the Magnetic Field in Solenoid

To apply Ampere’s law, consider an imaginary amperian loop in the shape of a rectangle \(abcd,\) as shown in the below figure. Along \(cd,\) the \(\vec B \cdot d\vec l\) is zero because the magnetic field is zero as it is outside the ideal solenoid. Along transverse sections \(bc\) and \(ad,\) the field will be perpendicular to the path so \(\vec B \cdot d\vec l\) will be zero. Thus, these two sections do not contribute. Now on path \(ab,\) the field is in the same direction as \(d\vec l.\) Let the field along \(ab\) be \(\vec B.\) Thus, the considered length of the Amperian loop is \(h(ab).\) And the total number of turns in the loop will be \(nh,\) where \(n\) be the number of turns per unit length.
The integral over the closed amperian rectangular path will be,
\(\oint {\vec B} \cdot d\vec l = Bh…..\left( 1 \right)\)
And the total current that passes through the area bounded by the path of integration is present in the right side of Ampere’s law equation. In this case,
\({I_{{\text{enc}}}} = \left( {{\text{number}}\,{\text{of}}\,{\text{turns}}\,{\text{inside}}\,{\text{the}}\,{\text{area}}} \right) \times \left( {{\text{current}}\,{\text{through}}\,{\text{each}}\,{\text{turn}}} \right)\)
\({I_{{\text{enc }}}} = \left({nh}\right)\left( I \right)\quad \ldots …\left( 2 \right)\)
Using Ampere’s law we have,
\(\oint {\vec B} \cdot d\vec l = {\mu _0}{I_{{\text{enc}}}}\)
Or,
Using equations \(\left( 1 \right)\) and \(\left( 2 \right),\) we have
\(Bh = {\mu _0}{I_{{\text{enc}}}},\)
\( \Rightarrow Bh = {\mu _0}I\left({nh} \right),\)
\( \Rightarrow B = {\mu _0}nI\quad \ldots \left( 3\right)\)
The above formula of magnetic field gives a good result when the length of the solenoid is much larger than the radius of the loop. This equation is used to obtain the magnitude of the magnetic field inside a long solenoid.
So, we can say that the magnetic field inside a long ideal solenoid depends on three main factors.
(i) Number of turns per unit length in the solenoid \(\left( n \right)\)
(ii) Strength of the current in the coil of the solenoid \(\left( I \right)\)
(iii) Permeability of the material inside the solenoid \(\left( \mu \right)\)

Solved Examples on Magnetic Field in a Solenoid

Q.1. What is the magnitude of the magnetic field inside a long solenoid if \(5\;\rm{A}\) of current is passing through it? It is given that the length of the solenoid is \(0.5\,\rm{m}\) and it has \(500\) turns.
Ans: The number of turns per unit length is,
\(n = \frac{{500}}{{0.5}} = 1000\,{\text{turns}}\,{\text{per}}\,{\text{meter}}\)
Hence, we can use the long solenoid formula.
\( \Rightarrow B ={\mu _0}nI\)
\( \Rightarrow B = 4\pi \times {10^{ – 7}} \times 1000 \times 5\;\rm{T}\)
\( \Rightarrow B = 6.28 \times {10^{ – 3}}\;\rm{T}\)

Q 2. A closed curve encircles several conductors. The line integral \(\int {\vec B} \cdot d\vec l\) around this curve is \(3.83 \times {10^{ – 7}}\;\rm{T – m}.\) Find the net current in the conductors?
Ans: (a) According to the Ampere Circuital Law, we know that
\(\int {\vec B} \cdot d\vec l = {\mu _0}{i_{{\text{net }}}}\)
\( \Rightarrow {i_{{\text{net }}}} = \frac{{\int {\vec B} \cdot d\vec l}}{{{\mu _0}}} = \frac{{3.83 \times {{10}^{ – 7}}}}{{4\pi \times {{10}^{ – 7}}}} = 0.3\;\rm{A}\)

Summary

A solenoid is a long wire that is wounded in the form of a helix that can produce a reasonably uniform magnetic field. The magnetic field is nearly uniform inside the solenoid and close to zero outside the solenoid. We use Ampere’s law to obtain an expression for the interior magnetic field in an ideal solenoid which states that the line integral of the magnetic field around a closed loop is proportional to the amount of current that is enclosed by the loop.

The magnitude of the magnetic field inside an ideal solenoid calculated from Ampere’s law is given as \(B = {\mu _0}nI.\) Where \(n\) is the number of turns per unit length, I is the current passing through it and \({\mu _0}\) is the constant of proportionality called the permeability of space and has a value of \(({\mu _0} = 4\pi \times {10^{ – 7}}\;\rm{T – m/A}).\) The magnitude of the magnetic field will be constant inside a long solenoid.

FAQs on Magnetic Field in a Solenoid

Q.1. What is the formula of the magnetic field of a solenoid?
Ans: The magnetic field inside an ideal solenoid has a magnitude of \(B = {\mu _0}nI.\) Here the number of turns per unit length is \(n,\) length is \(L,\) and current is \(I.\) It can be calculated from expression obtained from Ampere’s law.

Q.2. What will be the nature magnetic field inside an ideal solenoid?
Ans: Inside an ideal solenoid, the magnetic field is uniform, and field lines are straight parallel lines.

Q.3. What is the magnetic field outside a solenoid?
Ans: The magnetic field lines exist outside the solenoid, but the number of field lines per unit area outside the solenoid is much less as compared to the number of lines per unit area inside the solenoid. Hence the magnetic field is considered to be practically zero because the magnetic field outside the solenoid is so feeble. For an ideal solenoid, its value will be zero.

Q.4. What are the uses of the solenoid?
Ans: When a current passes through a solenoid, then it becomes an electromagnet. Electromagnets are used in magnetic door locks, water-pressure valves in air conditioning systems, MRI machines, hard disk drives, speakers, microphones, etc.

Q.5. What are the factors affecting solenoids?
Ans: Some of the important factors which affect the solenoid are given following:
(i) Number of turns per unit length in the solenoid
(ii) Strength of the current in the coil of the solenoid
(iii) Material inside the solenoid