A stationary magnetic dipole, m = m zˆ, is situated above an infinite uniform surface current, K = K xˆ (Fig. 12.44). (a) Find the torque on the dipole, using Eq. 6.1. (b) Suppose that the surface current consists of a uniform surface charge σ, moving at velocity v = υ xˆ, so that K = σv,

Question

A stationary magnetic dipole, [latex]m =m \hat{ z }[/latex], is situated above an infinite uniform surface current, [latex]K =K \hat{ x }[/latex] (Fig. 12.44).

(a) Find the torque on the dipole, using Eq. 6.1

N = m × B (6.1)

(b) Suppose that the surface current consists of a uniform surface charge σ, moving at velocity [latex]v =v \hat{ x }[/latex], so that K = σv, and the magnetic dipole consists of a uniform line charge λ, circulating at speed [latex]v ext { (same } v)[/latex] around a square loop of side l, as shown, so that [latex]m=\lambda v l^{2}[/latex]. Examine the same configuration from the point of view of system [latex]\overline{ S }[/latex], moving in the x direction at speed [latex]v[/latex]. In [latex]\overline{ S }[/latex], the surface charge is at rest, so it generates no magnetic field. Show that in this frame the current loop carries an electric dipole moment, and calculate the resulting torque, using Eq. 4.4.

N = p × E (4.4)

Accepted Answer

[latex] ext { (a) } B =-\frac{\mu_{0}}{2} K \hat{ y }( ext { Eq. } 5.58) ; N = m imes B ( ext { Eq. } 6.1), ext { so } N =-\frac{\mu_{0}}{2} m K(\hat{ z } imes \hat{ y })[/latex].

[latex]B = \begin{cases}+\left(\mu_{0} / 2 ight) K \hat{ y } & ext { for } z<0 \ -\left(\mu_{0} / 2 ight) K \hat{ y } & ext { for } z>0\end{cases}[/latex] (5.58)

[latex]N =\frac{\mu_{0}}{2} m K \hat{ x }=\frac{\mu_{0}}{2}\left(\lambda v l^{2} ight)(\sigma v) \hat{ x }=\frac{\mu_{0}}{2} \lambda \sigma v^{2} l^{2} \hat{ x }[/latex].

Charge density in the front side: [latex]\lambda_{0}\left(\lambda=\gamma \lambda_{0} ight)[/latex];

Charge density on the back side: [latex]\bar{\lambda}=\bar{\gamma} \lambda_{0}, ext { where } \bar{v}=\frac{2 v}{1+v^{2} / c^{2}}[/latex],

[latex] ext { so } \bar{\gamma}=\frac{1}{\sqrt{1-\frac{4 v^{2} / c^{2}}{\left(1+v^{2} / c^{2} ight)^{2}}}}=\frac{\left(1+v^{2} / c^{2} ight)}{\sqrt{1+2 \frac{v^{2}}{c^{2}}+\frac{v^{4}}{c^{4}}-4 \frac{v^{2}}{c^{2}}}}=\frac{1+v^{2} / c^{2}}{\sqrt{1-2 \frac{v^{2}}{c^{2}}+\frac{v^{4}}{c^{4}}}}=\frac{\left(1+v^{2} / c^{2} ight)}{\left(1-v^{2} / c^{2} ight)}=\gamma^{2}\left(1+\frac{v^{2}}{c^{2}} ight)[/latex]

Length of front and back sides in this frame: [latex]l / \gamma[/latex]. So net charge on back side is:

[latex]q_{+}=\bar{\lambda} \frac{l}{\gamma}=\gamma^{2}\left(1+\frac{v^{2}}{c^{2}} ight) \frac{\lambda}{\gamma} \frac{l}{\gamma}=\left(1+\frac{v^{2}}{c^{2}} ight) \lambda l[/latex]

Net charge on front side is:

[latex]q_{-}=\lambda_{0} \frac{l}{\gamma}=\frac{\lambda}{\gamma} \frac{l}{\gamma}=\frac{1}{\gamma^{2}} \lambda l[/latex]

So dipole moment (note: charges on sides are equal):

[latex]p =\left(q_{+} ight) \frac{l}{2} \hat{ y }-\left(q_{-} ight) \frac{l}{2} \hat{ y }=\left[\left(1+\frac{v^{2}}{c^{2}} ight) \lambda l \frac{l}{2}-\frac{1}{\gamma^{2}} \lambda l \frac{l}{2} ight] \hat{ y }=\frac{\lambda l^{2}}{2}\left(1+\frac{v^{2}}{c^{2}}-1+\frac{v^{2}}{c^{2}} ight) \hat{ y }=\frac{\lambda l^{2} v^{2}}{c^{2}} \hat{ y }[/latex].

[latex]E =\frac{\sigma_{0}}{2 \epsilon_{0}} \hat{ z }, ext { where } \sigma=\gamma \sigma_{0}, ext { so } N = p imes E =\frac{\lambda l^{2} v^{2}}{c^{2}} \frac{\sigma}{2 \epsilon_{0} \gamma}(\hat{ y } imes \hat{ z })=\frac{1}{\gamma} \frac{\mu_{0}}{2} \lambda \sigma l^{2} v^{2} \hat{ x }[/latex].

So apart from the relativistic factor of [latex]\gamma[/latex] the torque is the same in both systems — but in S it is the torque exerted by a magnetic field on a magnetic dipole, whereas in [latex]\overline{ S } [/latex] it is the torque exerted by an electric field on an electric dipole.

Question 12.65: A stationary magnetic dipole, m = m zˆ, is situated above an...

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