Compass needle in inhomogeneous magnetic field

Electricity and Magnetism Level 3

We consider a cylindrical coil of length l = 40 cm , l = 40\, \text{cm}, radius R = 4 cm , R = 4 \, \text{cm}, and number of turns N = 3200. N = 3200. A constant electric current I = 4 A I = 4 \, \text{A} flows through the coil, so that a magnetic field B ⃗ \vec B is created. Now we place a small compass inside the coil at z = − 1 2 l z = - \frac{1}{2} l . The compass needle is aligned along the magnetic field and has a magnetic moment of μ = 0.1 J / T \mu = 0.1 \, \text {J} / \text {T} .

What is the absolute value of the force F ⃗ \vec F exerted by the magnetic field on the compass needle?

Details: Along the z z -axis, the magnetic field of the cylindrical coil results in B ⃗ ( z ) = μ 0 N I 2 l [ z + l 2 R 2 + ( z + l 2 ) 2 − z − l 2 R 2 + ( z − l 2 ) 2 ] e ⃗ z , \vec B(z) = \frac{\mu_0 N I}{2 l} \left[ \frac{z + \frac{l}2}{\sqrt{R^2 + \left(z + \frac{l}2\right)^2}} - \frac{z - \frac{l}2}{\sqrt{R^2 + \left(z - \frac{l}2\right)^2}} \right] \vec e_z, where μ 0 = 4 π ⋅ 1 0 − 7 N / A 2 \mu_0 = 4 \pi \cdot 10^{-7} \,\text{N}/\text{A}^2 is the vacuum permeability

F ≈ 5 ⋅ 1 0 − 6 N F \approx 5 \cdot 10^{-6}\,\text{N} F ≈ 5 ⋅ 1 0 − 4 N F \approx 5 \cdot 10^{-4} \,\text{N} F ≈ 5 ⋅ 1 0 − 2 N F \approx 5 \cdot 10^{-2} \,\text{N} F ≈ 5 N F \approx 5 \,\text{N} F ≈ 5 ⋅ 1 0 2 N F \approx 5 \cdot 10^2 \,\text{N} The magnetic field exerts a pure torque on the compass needle so that F = 0 F = 0

Markus Michelmann by Markus Michelmann , 35, Germany

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1 solution

Inside the magnetic field of the coil, the compass needle has a potential energy E pot E_\text{pot} , that is given by the dot product of the magnetic moment μ ⃗ \vec \mu and the magnetic field B ⃗ \vec B : E pot = − μ ⃗ ⋅ B ⃗ ( z ) = − μ B ( z ) = − ∫ − ∞ z F ( z ′ ) d z ′ E_\text{pot} = - \vec \mu \cdot \vec B(z) = - \mu B(z) = - \int_{-\infty}^z F(z') dz' where F ( z ) F(z) is the corresponding magnetic force. Differentiation after z z yields F ( z ) = μ d B d z = μ 0 μ N I 2 l d d z [ z + l / 2 R 2 + ( z + l / 2 ) 2 − z − l / 2 R 2 + ( z − l / 2 ) 2 ] = μ 0 μ N I 2 l [ 1 R 2 + ( z + l / 2 ) 2 − 1 R 2 + ( z − l / 2 ) 2 − ( z + l / 2 ) 2 ( R 2 + ( z + l / 2 ) 2 ) 3 / 2 + ( z − l / 2 ) 2 ( R 2 + ( z − l / 2 ) 2 ) 3 / 2 ] = μ 0 μ N I 2 l [ R 2 ( R 2 + ( z + l / 2 ) 2 ) 3 / 2 − R 2 ( R 2 + ( z − l / 2 ) 2 ) 3 / 2 ] \begin{aligned} F(z) &= \mu \frac{dB}{dz} = \frac{\mu_0 \mu N I}{2 l} \frac{d}{dz}\left[ \frac{z + l/2}{\sqrt{R^2 + (z + l/2)^2}} - \frac{z - l/2}{\sqrt{R^2 + (z - l/2)^2}} \right] \\ &= \frac{\mu_0 \mu N I}{2 l} \left[ \frac{1}{\sqrt{R^2 + (z + l/2)^2}} - \frac{1}{\sqrt{R^2 + (z - l/2)^2}} - \frac{(z + l/2)^2}{(R^2 + (z + l/2)^2)^{3/2}} + \frac{(z - l/2)^2}{(R^2 + (z - l/2)^2)^{3/2}} \right]\\ &= \frac{\mu_0 \mu N I}{2 l} \left[ \frac{R^2}{(R^2 + (z + l/2)^2)^{3/2}} - \frac{R^2}{(R^2 + (z - l/2)^2)^{3/2}} \right] \end{aligned} Now we evaluate the force at z = − l / 2 z = -l/2 : F ( − l / 2 ) = μ 0 μ N I 2 l [ 1 R − R 2 ( R 2 + l 2 ) 3 / 2 ] ≈ R ≪ l μ 0 μ N I 2 l [ 1 R − R 2 l 3 ] ≈ R ≪ l μ 0 μ N I 2 l R = 4 π ⋅ 1 0 − 7 ⋅ 0.1 ⋅ 3200 ⋅ 4 2 ⋅ 0.4 ⋅ 0.04 N ≈ 0.05 N \begin{aligned} F(-l/2) &= \frac{\mu_0 \mu N I}{2 l} \left[ \frac{1}{R} - \frac{R^2}{(R^2 + l^2)^{3/2}} \right] \\ & \stackrel{R \ll l}{\approx} \frac{\mu_0 \mu N I}{2 l} \left[ \frac{1}{R} - \frac{R^2}{l^3} \right] \\ & \stackrel{R \ll l}{\approx} \frac{\mu_0 \mu N I}{2 l R} \\ & = \frac{4 \pi \cdot 10^{-7} \cdot 0.1 \cdot 3200 \cdot 4}{2 \cdot 0.4 \cdot 0.04} \,\text{N} \\ &\approx 0.05 \,\text{N} \end{aligned} (The approximations are optional, but simplify the numerical evaluation)

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