CBSE Class 12 Physics 2025: question 23(b)

Hints

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Step-by-step solution

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1. (i) $\vec{m}=I\vec{A}$, with magnitude $m=Ilb$. By the right-hand thumb rule, a clockwise current (as seen by the viewer) gives $\vec{m}$ perpendicular to the plane of the loop, pointing into the page (away from the viewer). In the diagram, draw the rectangle with clockwise arrows and a ⊗ at its centre marking $\vec{m}$.

   1 mark

2. (ii) Let the loop ABCD have arms AB and CD of length $b$ and arms BC and DA of length $l$. The rotation axis is perpendicular to $\vec{B}$, and the normal to the loop makes angle $\theta$ with $\vec{B}$. The forces on BC and DA are equal, opposite and act along the same line (the axis), so they cancel.

   ½ mark

3. Arms AB and CD are perpendicular to $\vec{B}$, so each experiences force $F=IbB$. Because their currents are opposite, one force is into and the other out of the plane. The vector sum of all forces is zero, so the net force is zero.

   ½ mark

4. The forces on AB and CD have different lines of action and form a couple. The perpendicular distance between them is $l\sin \theta$, so $\tau =F\cdot l\sin \theta =IbBl\sin \theta$.

   ½ mark

5. $\tau =I(lb)B\sin \theta =mB\sin \theta$. The torque turns $\vec{m}$ towards $\vec{B}$, so in vector form $\vec{\tau }=\vec{m}\times \vec{B}$.

   ½ mark

Answer: m = I l b, perpendicular to the loop (into the plane for a clockwise current); net force = 0; τ = IlbB sin θ, i.e. τ⃗ = m⃗ × B⃗.

Revise the chapter

• Chapter: Moving Charges and Magnetism (notes, formulas, where marks are lost and every past question)
• Topic: Torque on a Current Loop and Magnetic Dipole