Torque on a Bar Magnet in a Magnetic Field || Class 12 Physics Notes


Here we will study about the Torque on a Bar Magnet in a Magnetic Field and Important Points related to it.

When a bar magnet is placed in a uniform magnetic field the two poles experience a force. The forces are equal in magnitude and opposite in direction and do not have same line of action. They constitute a couple of forces which produces a torque. The torque tries to rotate the magnet so as to align it parallel to direction of field.

Net force on bar magnet = $mB \hat{ i }+ mB (-\hat{ i })=0$

Torque $\tau=$ force $\times$ perpendicular distance between forces

$\tau= MB (2 \ell \sin \theta)= MB \sin \theta$

In vector notation $\vec{\tau}=\vec{M} \times \vec{B}$

Important Points

  1. When a bar magnet is kept in a uniform magnetic field it experiences no force.
  2. The bar magnet experiences a torque $\vec{\tau}=\vec{M} \times \vec{B}$ The direction of torque is perpendicular to plane containing $\overrightarrow{ M }$ and $\overrightarrow{ B }$. This torque produces rotational motion of magnet.
  3. $\tau=\tau_{\max }= mB$when $\theta=\frac{\pi}{2}$ The magnet experiences maximum torque when dipole moment vector is perpendicular to magnetic field.
  4. $\tau=\tau_{\min }=0$ when $\theta=0$ or $\pi$. The magnet experiences minimum torque when dipole moment vector is parallel or antiparalled to magnetic field.

Also Read:

Biot Savart’s Law

 

Click here for the Video tutorials of Magnetic Effect of Current Class 12

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