Question
Derive an expression for the net torque on a rectangular current-carrying loop placed in a uniform magnetic field with its rotation axis perpendicular to the field.
✓
Answer
Consider a rectangular loop ABCD of length $I$, breadth $b$ and carrying a current $I$, placed in a uniform magnetic field of induction $\vec{B}$ with its rotation axis perpendicular to $\vec{B}$, from figure (a). To define the orientation of the loop in the magnetic field, we use a normal vector $\hat{n}$ that is perpendicular to the plane of the loop. The direction of $\hat{n}$ is given by a right hand rule: If the fingers of right hand are curled in the direction of current in the loop, the outstretched thumb is the direction of $\hat{n}$. Suppose the normal vector $\hat{n}$ of the loop makes an arbitrary angle with $\vec{B}$, as shown in figure (b).
In the side view, Fig. 10.15 (b), the sides CD, DA, AB and BC have been labelled as 1,2,3 and 4 , respectively. In this view, the current in side $1(C D)$ is out of the page as shown by a $\odot$ while that in side $3(A B)$ is into the page shown by a $\otimes$.
For side 2 (AD) and side 4 (BC), the length of the conductor $|\vec{L}|= b$ and the angle between $\vec{L}$ and $\vec{B}$ is $\left(90^{\circ}-\theta\right)$. Hence, the forces on sides 2 and 4 are equal in magnitude :
$
F_2=F_4=l b B \sin \left(90^{\circ}-\theta\right)= lbB \cos \theta
$
However, $\vec{F}_2$ is directed out of the page while $\vec{F}_4$ is into, and because their common line of action is through the centre of the loop, their net torque is zero. For side $1(C D)$ and side 3 (AB), $|\vec{L}|=1$ and $\vec{L}$ is perpendicular to $\vec{B}$. Hence, the forces $\vec{F}_1$, and $F \vec{F}_3$ have the same magnitude : $F_1=F_3=I I B$
But their lines of action being different, they constitute a couple.
Moment arm of the couple $=b \sin \theta$
$\therefore$ Torque exerted by the couple $=$ force of the couple $x$ moment arm of couple
$
\therefore \tau=(I \mid B)(b \sin \theta)
$
in the clockwise sense in figure (b). The torque tends to rotate the loop so as to align its normal vector $h$ with the direction of the magnetic field.
$
\therefore \tau=1( lb ) B \sin \theta= MB \sin \theta
$
where $A= lb$ is the area of the loop. For a rectangular coil of $N$ turns in place of a single-turn loop,
$
\tau= NIAB \sin \theta
$
This is the required expression for the net torque. The torque has maximum magnitude for $\theta$ $=90^{\circ}$, that is when $\hat{n}$ is perpendicular to $\vec{B}$ or, in other words, the plane of the coil is parallel to the field.
$
\tau_{\max }= NIAB
$
In the side view, Fig. 10.15 (b), the sides CD, DA, AB and BC have been labelled as 1,2,3 and 4 , respectively. In this view, the current in side $1(C D)$ is out of the page as shown by a $\odot$ while that in side $3(A B)$ is into the page shown by a $\otimes$.For side 2 (AD) and side 4 (BC), the length of the conductor $|\vec{L}|= b$ and the angle between $\vec{L}$ and $\vec{B}$ is $\left(90^{\circ}-\theta\right)$. Hence, the forces on sides 2 and 4 are equal in magnitude :
$
F_2=F_4=l b B \sin \left(90^{\circ}-\theta\right)= lbB \cos \theta
$
However, $\vec{F}_2$ is directed out of the page while $\vec{F}_4$ is into, and because their common line of action is through the centre of the loop, their net torque is zero. For side $1(C D)$ and side 3 (AB), $|\vec{L}|=1$ and $\vec{L}$ is perpendicular to $\vec{B}$. Hence, the forces $\vec{F}_1$, and $F \vec{F}_3$ have the same magnitude : $F_1=F_3=I I B$
But their lines of action being different, they constitute a couple.
Moment arm of the couple $=b \sin \theta$
$\therefore$ Torque exerted by the couple $=$ force of the couple $x$ moment arm of couple
$
\therefore \tau=(I \mid B)(b \sin \theta)
$
in the clockwise sense in figure (b). The torque tends to rotate the loop so as to align its normal vector $h$ with the direction of the magnetic field.
$
\therefore \tau=1( lb ) B \sin \theta= MB \sin \theta
$
where $A= lb$ is the area of the loop. For a rectangular coil of $N$ turns in place of a single-turn loop,
$
\tau= NIAB \sin \theta
$
This is the required expression for the net torque. The torque has maximum magnitude for $\theta$ $=90^{\circ}$, that is when $\hat{n}$ is perpendicular to $\vec{B}$ or, in other words, the plane of the coil is parallel to the field.
$
\tau_{\max }= NIAB
$
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