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Chapter 8: Problem 12

Two circular wire rings are parallel to each other, share the same axis, are of radius \(a\), and are separated by distance \(d\), where \(d<

Short Answer

Expert verified
Answer: The approximate force of attraction between the two wire rings is given by \(F = \frac{\mu_0 I^2 a^2 d}{(a^2 + d^2)^2}\). The relative orientation of the currents should be in the same direction (either clockwise or counterclockwise for both rings) for the rings to attract each other.

Step by step solution

01

Determine the magnetic field produced by a circular wire ring

Given that each wire ring has a radius "a" and carries a current "I", we can use Ampere's Law to determine the magnetic field produced by one ring. The magnetic field at a distance "d" above the center of the ring can be determined using the equation: \(B = \frac{\mu_0 I a^2}{2(a^2 + d^2)^{3/2}}\)
02

Determine the force between the two rings

The force between the two rings can be determined using the equation: \(F = I_1 I_2 \oint_{C_1} \oint_{C_2} \frac{\mu_0 d \vec{dl_1} \cdot \vec{dl_2}}{4 \pi (a^2 + d^2)^2}\) Since both rings have the same current "I" and we are assuming a constant magnetic field, we can simplify the equation as: \(F = I^2 \oint_{C_1} \oint_{C_2} \frac{\mu_0 d \vec{dl_1} \cdot \vec{dl_2}}{4 \pi (a^2 + d^2)^2}\)
03

Evaluate the double integral

To find the force, we need to evaluate the double integral in the equation: \(F = I^2 \oint_{C_1} \oint_{C_2} \frac{\mu_0 d \vec{dl_1} \cdot \vec{dl_2}}{4 \pi (a^2 + d^2)^2} = I^2 \mu_0 d \oint_{C_1} \oint_{C_2} \frac{ \vec{dl_1} \cdot \vec{dl_2}}{4 \pi (a^2 + d^2)^2}\) Since the rings are parallel to each other, we can use their symmetry to further simplify this expression: \(F = I^2 \mu_0 d \frac{L_1 L_2}{4 \pi (a^2 + d^2)^2}\) Where \(L_1\) and \(L_2\) represent the lengths of the perimeters of the rings. Since both rings have the same radius, their perimeters are equal, and we can write: \(F = I^2 \mu_0 d \frac{2 \pi a \cdot 2 \pi a}{4 \pi (a^2 + d^2)^2}\)
04

Compute the force of attraction

Simplify the expression and find the force of attraction between the two rings: \(F = I^2 \mu_0 d \frac{4 \pi^2 a^2}{4 \pi (a^2 + d^2)^2} = \frac{\mu_0 I^2 a^2 d}{(a^2 + d^2)^2}\)
05

Determine the relative orientations of the currents

To attract each other, the circular wire rings should have currents flowing in the same direction. Their magnetic fields will interact, leading to an attractive force between the rings. Thus, the relative orientation of the currents should be in the same direction (either clockwise or counterclockwise for both rings) for them to experience an attraction force. In conclusion, the approximate force of attraction between the two wire rings is given by: \(F = \frac{\mu_0 I^2 a^2 d}{(a^2 + d^2)^2}\) And the relative orientation of the currents should be in the same direction for the rings to attract each other.

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Most popular questions from this chapter

A current of 6 A flows from \(M(2,0,5)\) to \(N(5,0,5)\) in a straight, solid conductor in free space. An infinite current filament lies along the \(z\) axis and carries \(50 \mathrm{~A}\) in the \(\mathbf{a}_{z}\) direction. Compute the vector torque on the wire segment using an origin at: \((a)(0,0,5) ;(b)(0,0,0) ;(c)(3,0,0)\).

A rectangular loop of wire in free space joins point \(A(1,0,1)\) to point \(B(3,0,1)\) to point \(C(3,0,4)\) to point \(D(1,0,4)\) to point \(A\). The wire carries a current of \(6 \mathrm{~mA}\), flowing in the \(\mathbf{a}_{z}\) direction from \(B\) to \(C\). A filamentary current of 15 A flows along the entire \(z\) axis in the \(\mathbf{a}_{z}\) direction. \((a)\) Find \(\mathbf{F}\) on side \(B C\). \((b)\) Find \(\mathbf{F}\) on side \(A B\). \((c)\) Find \(\mathbf{F}_{\text {total }}\) on the loop.

Uniform current sheets are located in free space as follows: \(8 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) at \(y=0,-4 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) at \(y=1\), and \(-4 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) at \(y=-1\). Find the vector force per meter length exerted on a current filament carrying \(7 \mathrm{~mA}\) in the \(\mathbf{a}_{L}\) direction if the filament is located at \((a) x=0, y=0.5\), and \(\mathbf{a}_{L}=\mathbf{a}_{z} ;\) (b) \(y=0.5, z=0\), and \(\mathbf{a}_{L}=\mathbf{a}_{x} ;(c) x=0, y=1.5\), and \(\mathbf{a}_{L}=\mathbf{a}_{z}\)

Calculate values for \(H_{\phi}, B_{\phi}\), and \(M_{\phi}\) at \(\rho=c\) for a coaxial cable with \(a=2.5 \mathrm{~mm}\) and \(b=6 \mathrm{~mm}\) if it carries a current \(I=12 \mathrm{~A}\) in the center conductor, and \(\mu=3 \mu \mathrm{H} / \mathrm{m}\) for \(2.5 \mathrm{~mm}<\rho<3.5 \mathrm{~mm}, \mu=5 \mu \mathrm{H} / \mathrm{m}\) for \(3.5 \mathrm{~mm}<\rho<4.5 \mathrm{~mm}\), and \(\mu=10 \mu \mathrm{H} / \mathrm{m}\) for \(4.5 \mathrm{~mm}<\rho<6 \mathrm{~mm}\). Use \(c=:(a) 3 \mathrm{~mm} ;(b) 4 \mathrm{~mm} ;(c) 5 \mathrm{~mm} .\)

Conducting planes in air at \(z=0\) and \(z=d\) carry surface currents of \(\pm K_{0} \mathbf{a}_{x} \mathrm{~A} / \mathrm{m} .(a)\) Find the energy stored in the magnetic field per unit length \((0
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