[NOTES/EM-07005]-Torque on a Current Carrying Loop

An expression for torque on a current carrying loop in magnetic field is obtained. It is shown that the torque is given by \(\vec{\tau}=\vec{m}\times\vec{B}\), where \(vec{m}\) is the magnetic moment of the loop, \(|\vec{m}|=\) current \(\times\) area.

Torque on a Current Carrying Loop

Suppose we have a current carrying loop placed in a uniform magnetic field. It will experience no force. We will no show that the torque on the loop is nonzero. The force and torque experienced by a line element is \begin{align} d\vec{F} = & Id\vec{r}\times\vec{B}\label{eq001}\\ d\vec{\tau} = & I\vec{r}\times(d\vec{r}\times \vec{B})\label{eq002} \end{align}

QFY 1:Show that the net force on a current loop obtained by integrating \eqRef{eq001} around a closed loop is zero. The net torque on a current loop is obtained by integrating eq002 over the loop. \begin{align} \vec{\tau} = & I\oint \vec{r}\times(d\vec{r}\times\vec{B})\label{eq003}\\ =& I\oint\left((\vec{r}\cdot\vec{B})d\vec{r}-(\vec{r}\cdot d\vec{r})\cdot\vec{B}\right)\label{eq004}\\ =& I\oint(\vec{r}\cdot\vec{B})d\vec{r}-I\vec{B}\oint (\vec{r}\cdot d\vec{r})\label{eq005}\\ =&I\oint (\vec{r}\cdot\vec{B})d\vec{r}\label{eq006} \end{align} we assumed that the magnetic field is uniform we have used the result that $\oint \vec{r}\cdot\vec{r}=0$. We want to put the answer for the torque in a nice form. If $\phi(\vec{r})$ is function of $\vec{r}$, we note that \begin{align} \int_1^2 d\phi(\vec{r}) =&\phi(\vec{r}_2)-\phi(\vec{r}_2) \label{eq007}\\ \therefore\qquad \oint d\phi(\vec{r}) = & 0\label{eq008} \end{align} Taking $\phi=(\vec{r}\cdot\vec{B})\vec{r}$ and noting that \begin{align} d\phi = d\big[(\vec{r}\cdot\vec{B})\vec{r}\big]=(\vec{B}\cdot d\vec{r}) \vec{r}+(\vec{B}\cdot\vec{r})d\vec{r}\label{eq009} \end{align} Using \eqRef{eq009} in \EqRef{eq008} we get \begin{align} \oint d\big((\vec{r}\cdot\vec{B})\vec{r}\big) = & 0\label{eq010}\\ \Rightarrow\quad \oint(\vec{r}\cdot\vec{B})d\vec{r}+\oint (\vec{B}\cdot d\vec{r})\vec{r} = & 0\label{eq011} \end{align} Multiplying \eqRef{eq011} by $-\frac{1}{2}$ and adding to the expression \eqRef{eq006} for torque, we get \begin{align} \vec{\tau} = & I\oint(\vec{r}\cdot\vec{B})d\vec{r}\label{eq012}\\ = & \frac{I}{2}\left(\oint(\vec{r}\cdot\vec{B})d\vec{r} - \oint(\vec{B}\cdot d\vec{r})\vec{r}\right)\label{eq013}\\ = & \frac{I}{2}\oint(\vec{r}\times d\vec{r})\times\vec{B}\label{eq014}\\ = & \vec{m} \times \vec{B}\label{eq015} \end{align} where $\vec{m}$, \begin{align} \boxed{\vec{m} = \frac{I}{2} \oint\vec{r}\times d\vec{r}}\,\,,\label{eq016} \end{align} is known as the \textbf{magnetic moment} of the current loop and for a planar loop it is equal to \begin{align} \boxed{\vec{m} = IA\hat{n}}\label{eq017} \end{align} where $A$ is the area of the loop and $\hat{n}$ is unit vector normal to the plane of the loop.

QFY 2 : Show that for a closed loop $\gamma$ the second integral, $\oint_\gamma(\vec{r}\cdot d\vec{r})$, in eq 005 vanishes.

QFY 3  : Show that for a planar loop the magnetic moment is given by eq017 

References

• Sec 33-4 Torque on a current loop David Halliday and R. Resnick, Physics Vol-II, Second Edition, (Revised Printing 1966), New Age International Publishers, New Delhi,(1966).