Question
Write the formula for the torque acting on a current$-$carrying coil placed in uniform magnetic field and explain its stable and unstable equilibrium states.
✓
Answer
$\rightarrow$ Torque exerted on a current carrying coil placed in a uniform magnetic field.
$\vec{\tau}=\vec{m} \times \vec{B}$
$\therefore \tau=m B \sin \theta$
$\rightarrow$ When, $\vec{m}$ and $\overrightarrow{ B }$ are parallel $\left(\theta=0^{\circ}\right)$ then $\tau=0$ this state of coil is called the stable equilibrium.
$\rightarrow$ In this condition, if the coil is rotated by a small angle the coil produces a torque which brings it back to its original position.
$\rightarrow$ When $\vec{m}$ and $\vec{B}$ are anti parallel $(\theta=\pi)$, then $\tau=0$ then it is called the unstable equilibrium state.
$\rightarrow$ In this condition, if the coil is rotated by small angle the coil produces a torque which can not bring it back to its original position.
$\vec{\tau}=\vec{m} \times \vec{B}$
$\therefore \tau=m B \sin \theta$
$\rightarrow$ When, $\vec{m}$ and $\overrightarrow{ B }$ are parallel $\left(\theta=0^{\circ}\right)$ then $\tau=0$ this state of coil is called the stable equilibrium.
$\rightarrow$ In this condition, if the coil is rotated by a small angle the coil produces a torque which brings it back to its original position.
$\rightarrow$ When $\vec{m}$ and $\vec{B}$ are anti parallel $(\theta=\pi)$, then $\tau=0$ then it is called the unstable equilibrium state.
$\rightarrow$ In this condition, if the coil is rotated by small angle the coil produces a torque which can not bring it back to its original position.
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