Class 12 > Unit # 18:Magnetic Fields > Ampere's Law & Magnetic Field Inside a Solenoid

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Electromagnetism - Ampere's Law and Solenoids

Q. State and explain Ampere’s Law.

Ampere's Law: Statement

"The sum of the product of the tangential component of magnetic field of induction and the length of an element of a closed curve taken in the magnetic field is $\mu_0$ times the current which passes through the area bounded by the curve."

Figure: Closed circular path of radius $r$ around a long straight current-carrying conductor.

Explanation and Derivation

Consider a long straight wire carrying a current $I$. The lines of force are concentric circles with their common center on the wire. Consider a circle of radius $r$ from these circles. The magnetic field at all points on this circle is of the same magnitude.

Biot and Savart experimentally found that the magnitude of the field depends directly on the current $I$ and inversely on the distance $r$ from the conductor:

$B \propto \frac{I}{r} \implies B = \frac{\mu_0 I}{2\pi r} \quad \text{--- (1)}$

Where $\mu_0$ is called the permeability of free space, and its value is $\mu_0 = 4\pi \times 10^{-7} \, \text{Wb/(A}\cdot\text{m)}$.

Let the circular path be divided into small elements each of length $\Delta l$. Since $\Delta l$ is very small, the tangential component of the field $B$ and $\Delta l$ remain parallel ($\theta = 0^\circ$). Thus, for a single element:

$B \cdot \Delta l = B \Delta l \cos(0^\circ) = B \Delta l$

The sum of these products for all elements around the closed loop is:

$\sum (B \cdot \Delta l) = B \sum \Delta l$

Since $\sum \Delta l = 2\pi r$ (the circumference of the circular path), we get:

$\sum (B \cdot \Delta l) = B (2\pi r) \quad \text{--- (2)}$

Substituting the value of $B$ from equation (1) into equation (2):

$\sum (B \cdot \Delta l) = \left(\frac{\mu_0 I}{2\pi r}\right) (2\pi r) = \mu_0 I$

$\sum (B \cdot \Delta l) = \mu_0 I \quad \text{(Ampere's Circuital Law)}$

Thus, Ampere's law can also be stated as: "The sum of the dot product of magnetic field of induction and length of an element of a closed path in the magnetic field is $\mu_0$ times the current enclosed by the path."

Q. Using Ampere’s Law derive an expression for the magnetic field of induction inside a long current carrying solenoid.

Solenoid Definition

"A solenoid is a coil of insulated copper wire wound in close turns on a cylindrical former. When current passes through it, a magnetic field is produced which is uniform and strong inside the solenoid, while outside it the field is negligibly weak."

Figure: Rectangular closed path $abcda$ inside and outside a current-carrying solenoid.

Derivation of Magnetic Field Inside a Solenoid

Consider a solenoid through which current $I$ is passing. To determine the magnetic field of induction $B$ at any point inside the solenoid, imagine a closed rectangular path $abcda$ divided into four elements of length $\Delta l_1, \Delta l_2, \Delta l_3,$ and $\Delta l_4$:

1. Along side 1 (inside the solenoid): $\Delta l_1$ is parallel to $\vec{B}$, so:
   $\sum B \Delta l_1 = B L$
2. Along side 2 (outside the solenoid): The magnetic field is very weak ($B = 0$), so:
   $\sum B \Delta l_2 = 0$
3. Along sides 3 and 4 (perpendicular to field): $\Delta l_3$ and $\Delta l_4$ are perpendicular to $\vec{B}$ ($\theta = 90^\circ$), so:
   $\sum B \Delta l_3 = 0 \quad \text{and} \quad \sum B \Delta l_4 = 0$

The total sum of all dot products around the closed rectangular path $abcda$ is:

$\oint B \cdot \Delta l = BL + 0 + 0 + 0 = BL \quad \text{--- (i)}$

According to Ampere’s Circuital Law:

$\oint B \cdot \Delta l = \mu_0 \times (\text{Total Current Enclosed})$

If there are $n$ number of turns per unit length of the solenoid, and each turn carries a current $I$, the total number of turns in length $L$ is $nL$, and the enclosed current is $nLI$:

$\oint B \cdot \Delta l = \mu_0 (n L I) \quad \text{--- (ii)}$

Comparing equation (i) and equation (ii), we get:

$BL = \mu_0 n L I$

Dividing both sides by $L$ yields the final expression for the magnetic field inside a long solenoid:

$B = \mu_0 n I \quad \text{(Proved)}$

Conclusion:
The magnitude of the magnetic field inside a solenoid depends directly on the permeability of free space ($\mu_0$), the number of turns per unit length ($n$), and the current ($I$). The direction of the field is directed along the geometric axis of the solenoid.

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