Physics Lesson 16.6.5 - Ampere's Law Applied in Solenoids and Toroids

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Welcome to our Physics lesson on Ampere's Law Applied in Solenoids and Toroids, this is the fifth lesson of our suite of physics lessons covering the topic of Ampere's Law, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Ampere's Law Applied in Solenoids and Toroids

In tutorial 16.2 we have explained that the magnetic field produced by a solenoid which contains N turns is

B = μ₀ ∙ N ∙ I/L

If we express this equation in terms of the number of turns per unit length n instead of N/L, we obtain for the magnetic field inside the solenoid

B = μ₀ ∙ n ∙ I

This formula is obtained even when we use Ampere's law approach. For this, we have to consider a rectangular Amperial loop abcd as shown below.

Physics Tutorials: This image provides visual information for the physics tutorial Ampere's Law

An ideal solenoid has a uniform magnetic field inside (the parallel lines in the figure) and zero outside it. Using the rectangular loop abcda, we write the closed integral ∮BdL as the sum of four integrals (one for each segment). Thus, we have:

∮BdL = ∫^b_aBdL + ∫^c_bBdL + ∫^d_cBdL + ∫^a_dBdL

The only integral, which has a non-zero value, is the first one. On the other hand, the integral according the path cd is zero because magnetic field outside the solenoid is zero. Also, the integrals according the paths bc and da are both zero as the magnetic field lines are perpendicular to the paths (cos 90° = 0). Therefore, if we denote the length ab = h we obtain

∮BdL = B ∙ h

As for the net current I_encl enclosed within the Amperian loop, we have

i_encl = I ∙ (n ∙ h)

where 'I' is the current flowing through the entire solenoid.

Combining the general form of Ampere's law

∮BdL = μ₀ ∙ i_encl

with the above (transformed) form of this law, we obtain

B ∙ h = μ₀ ∙ I ∙ (n ∙ h)

B = μ₀ ∙ n ∙ I

This is the same formula obtained in 16.2 for the magnetic field inside a solenoid but now using Ampere's law.

A toroid is a hollow solenoid that has been curved until its two ends meet, forming a sort of hollow bracelet.

We can determine the magnetic field inside the toroid (inside the bracelet-like shape) using Ampere's law.

From symmetry of toroid it is easy to conclude that the magnetic field lines inside a toroid are concentric circles. We can choose a concentric circle of radius r as an Amperian loop.

When solving the integral of Ampere's Law for toroid

∮BdL = μ₀ ∙ i_encl

we obtain

B ∙ 2π ∙ r = μ₀ ∙ i_encl ∙ N

where i is the current in the toroid windings (it is positive for windings included inside the Amperian loop) and N is the total number of turns. Hence, we obtain for the magnetic field inside the toroid:

B = μ₀ ∙ i_encl ∙ N/2π ∙ r

Unlike in solenoids, the magnetic field B is not constant throughout the cross-section of toroids. In addition, the magnetic field outside a toroid is zero.

The direction of magnetic field in toroid can be determined in the same way as in solenoids, i.e. by using the curled right-hand rule in which we grasp the toroid with our right hand (the four fingers are in the direction of current) and the outstretched thumb shows the direction of magnetic field.

Example 2

A toroid having 500 turns has an inner diameter d = 3 cm and an outer diameter D = 5 cm. What is the magnetic field at centre of cross section of toroid if the current flowing through it is 20 mA?

Solution 2

The centre of cross-section of toroid is at any midpoint between the inner and outer radius. This position corresponds to the value of r in the formula (as in the figure shown in theory). Thus, we have

r = d + D/2
= 3 cm + 5cm/2
= 4 cm
= 4 × 10^-2 m

Giving that N = 500 = 5 × 10² and i_encl = 20 mA = 0.02 A = 2 × 10^-2 A, we obtain

B = μ₀ ∙ i_encl ∙ N/2π ∙ r
= (4π × 10^-7 N/A² ) ∙ (2 × 10^-2 A) ∙ (5 × 10² )/2π ∙ (4 × 10^-2 m)
= 5 × 10^-5 T

You have reached the end of Physics lesson 16.6.5 Ampere's Law Applied in Solenoids and Toroids. There are 5 lessons in this physics tutorial covering Ampere's Law, you can access all the lessons from this tutorial below.

More Ampere's Law Lessons and Learning Resources

Magnetism Learning Material
Tutorial ID	Physics Tutorial Title	Revision Notes	Revision Questions
16.6	Ampere's Law
Lesson ID	Physics Lesson Title	Lesson	Video Lesson
16.6.1	Useful Mathematical Background: Integral of a function - Geometrical approach
16.6.2	Ampere's Law
16.6.3	Magnetic Field Outside a Long Straight Wire with Current
16.6.4	Magnetic Field Inside a Long Straight Wire with Current
16.6.5	Ampere's Law Applied in Solenoids and Toroids

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