Question
Derive the equation for the torque acting on a current carrying loop placed at an angle $\theta$ with uniform magnetic field $(\vec{B})$.
✓
Answer
→As shown in the figure (a), the coil is placed in such a way that the magnetic field marks an angle $\theta$ with normal to the coil.
→The current passing through the coil is I. The length and width of the coil are $a$ and $b$ respectively.
→The force on the arm BC of the coil due to the current is $F _1= I a B \cos \theta$ and the force on the $\operatorname{arm} AD$ is $F _2= I a B \cos \theta$.
→Since, both these forces are equal, opposite and act along the axis of the coil, the resultant force and torque is zero.
→Similarly, the force acting perpendicular to the side AB and CD of the coil is $F _1{ }^{\prime}= F _2{ }^{\prime}= I b B$ respectively.
→Since, these two forces are equal in magnitude and opposite to each other the net force is zero. But they are non collinear, they form a couple, so a torque is exerted on the coil.
→From the figure (b) the torque acting on the coil.
$\begin{aligned}
\tau & =\left( F _1{ }^{\prime}\right) \frac{a}{2} \sin \theta+\left( F _2{ }^{\prime}\right)\left(\frac{a}{2}\right) \sin \theta \\
\therefore \tau & = I b B\left(\frac{a}{2}\right) \sin \theta+ I b B\left(\frac{a}{2}\right) \sin \theta \\
\therefore \quad \tau & =2 I b B \frac{a}{2} \sin \theta \\
\therefore \tau & = IB (a b) \sin \theta
\end{aligned}$
But $a b= A$ is the area of the rectangle.
$\therefore \tau= IAB \sin \theta$
→Vector form, $\vec{\tau}=I \vec{A} \times \vec{B}$
→But the magnetic dipole moment $\vec{m}= I \overrightarrow{ A }$
→The direction of the magnetic dipole moment is in the direction of area vector.
→Thus, torque acting on the coil $\vec{\tau}=\vec{m} \times \overrightarrow{ B } \ldots$
→If there are N turns in the coil then the torque is $\tau= NIAB \sin \theta$
Where, NIA $=m$ (magnetic moment)
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