Question 5.23: A small circular loop of radius a is centered at the origin ......
A small circular loop of radius a is centered at the origin in the xy–plane, carrying a current I in the counterclockwise direction in a uniform magnetic flux density \pmb{B}= B_{y} \pmb{a}_{y}+ B_{z} \pmb{a}_{z}, as shown in Fig. 5.40. Find the torque on the loop about the origin.
From Eq. (5-135b), the differential magnetic force on a differential current element I d\pmb{l}= I a d\phi \pmb{a}_{\phi } is
\boxed{\pmb{F}_{m} =I \oint_{C}{d\pmb{l}\times \pmb{B}}} [N] (5-135b)
d\pmb{F} = I a d\phi \pmb{a}_{\phi }\times (B_{y} \pmb{a}_{y}+ B_{z} \pmb{a}_{z})\\ \quad \quad = – a I B_{y}\sin \phi d\phi \pmb{a}_{z} + aIB_{z}d\phi \pmb{a}_{\rho } \\ \quad \quad \equiv \pmb{F}_{1} +\pmb{F}_{2}The differentia torque about the origin due to Idl is
d\pmb{T} =\pmb{r} \times (\pmb{F}_{1} +\pmb{F}_{2}) = a \pmb{a}_{\rho } \times (- a I B_{y} \sin \phi d\phi \pmb{a}_{z } + a I B_{z} d \phi \pmb{a}_{\rho })\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad = a^{2} I B_{y} \sin \phi d\phi \pmb{a}_{\phi }The net torque on the loop is
\pmb{T}=\int{d\pmb{T}} = a^{2} I B_{y} \int_{\phi =0}^{\phi =2\pi }{ \pmb{a}_{\phi } \sin \phi d\phi } (5-151)
The unit vector \pmb{a}_{\phi } cannot be taken outside the integral sign, because it is a function of Φ. Using the relation \pmb{a}_{\phi } = -\sin \phi \pmb{a}_{x} + \cos \phi \pmb{a}_{y}, Eq. (5-151) is rewritten as
\pmb{T}= a^{2} I B_{y} \int_{\phi =0}^{\phi =2\pi }{ (-\sin ^{2}\phi \pmb{a}_{x} + \sin \phi \cos \phi \pmb{a}_{y}) d\phi } \\ \quad =- \pi a^{2}IB_{y}\pmb{a}_{x}With the aid of the magnetic dipole moment \pmb{m}= \pi a^{2}I\pmb{a}_{z}, the torque on the loop is, in vector notation,
\pmb{T = m \times B} (5-152)
We see from Eqs. (5-150) and (5-152) that the torque on a current-carrying loop has the same form, in vector notation, regardless of the shape of the loop.
\boxed{\pmb{T = m \times B}} [N m]⋅ (5-150)