Question
Derive an expression for the net torque on a rectangular current carrying loop placed in a uniform magnetic field with its rotational axis perpendicular to the field.
✓
Answer
i. Consider rectangular loop abcd placed in a uniform magnetic field $\overrightarrow{ B }$ such that the sides ab and $c d$ are perpendicular to the magnetic field $\overrightarrow{ B }$ but the sides bc and da are not, as shown in figure (a) below:
Fig (a): Loop abcd placed in a uniform
ii. The force $\overrightarrow{ F }_4$ on side 4 (bc) will be $\overrightarrow{ F }_4= II _2 B \sin \left(90^{\circ}\right.$ $\theta)$
iii. The force $\overrightarrow{ F }_2$ on side 2 (da) will be equal and opposite to $\overrightarrow{ F }_4$ and both act along the same line. Thus, $\overrightarrow{ F }_2$ and $\overrightarrow{ F }_4$ will cancel out each other.
iv. The magnitudes of forces $\overrightarrow{ F }_1$ and $\overrightarrow{ F }_3$ on sides 1 (cd) and $3( ab )$ will be $\|_1 B \sin 90^{\circ}$ i.e., $\|_1 B$. These two forces do not act along the same line and hence they produce a net torque.
v. This torque results into rotation of the loop so that the loop is perpendicular to the direction of $\vec{B}$, the magnetic field.
Fig (b): Side view of the loop abcd at an angle $\theta$ vi. Now the moment arm is $\frac{1}{2}\left(l_2 \sin \theta\right)$ about the central axis of the loop. Hence, the torque $\tau$ due to forces $\overrightarrow{ F }_1$ and $\overrightarrow{ F }_3$ will be
$ \tau=\left( Il _1 B \frac{1}{2} l _2 \sin \theta\right)+\left( Il _1 B \frac{1}{2} l _2 \sin \theta\right)$
$= I l_1 l _2 B \sin \theta $
vii. If the current-carrying loop is made up of multiple turns $N$, in the form of a flat coil, the total torque will be
$\tau^{\prime}= N \tau= NIl _1 l _2 B \sin \theta$
$\tau^{\prime}=(N I A) B \sin \theta$; where $A$ is the are enclosed by the coil $=I_1 I_2$
This is the required expression.
Fig (a): Loop abcd placed in a uniform
ii. The force $\overrightarrow{ F }_4$ on side 4 (bc) will be $\overrightarrow{ F }_4= II _2 B \sin \left(90^{\circ}\right.$ $\theta)$
iii. The force $\overrightarrow{ F }_2$ on side 2 (da) will be equal and opposite to $\overrightarrow{ F }_4$ and both act along the same line. Thus, $\overrightarrow{ F }_2$ and $\overrightarrow{ F }_4$ will cancel out each other.
iv. The magnitudes of forces $\overrightarrow{ F }_1$ and $\overrightarrow{ F }_3$ on sides 1 (cd) and $3( ab )$ will be $\|_1 B \sin 90^{\circ}$ i.e., $\|_1 B$. These two forces do not act along the same line and hence they produce a net torque.
v. This torque results into rotation of the loop so that the loop is perpendicular to the direction of $\vec{B}$, the magnetic field.
Fig (b): Side view of the loop abcd at an angle $\theta$ vi. Now the moment arm is $\frac{1}{2}\left(l_2 \sin \theta\right)$ about the central axis of the loop. Hence, the torque $\tau$ due to forces $\overrightarrow{ F }_1$ and $\overrightarrow{ F }_3$ will be
$ \tau=\left( Il _1 B \frac{1}{2} l _2 \sin \theta\right)+\left( Il _1 B \frac{1}{2} l _2 \sin \theta\right)$
$= I l_1 l _2 B \sin \theta $
vii. If the current-carrying loop is made up of multiple turns $N$, in the form of a flat coil, the total torque will be
$\tau^{\prime}= N \tau= NIl _1 l _2 B \sin \theta$
$\tau^{\prime}=(N I A) B \sin \theta$; where $A$ is the are enclosed by the coil $=I_1 I_2$
This is the required expression.
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