Skip to content
Question of 55

Q.Derive an expression for magnetic force. How the direction of magnetic force is determined ?

Punjab PsebPSEB Punjab Class 12 Board 2025Subjective· 2mImportance★★★★★
0% · 0/55 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

A charge qq moving with velocity v⃗\vec{v} in magnetic field B⃗\vec{B} experiences a force F⃗=qv⃗×B⃗\vec{F} = q\vec{v}\times\vec{B}, always perpendicular to both v⃗\vec{v} and B⃗\vec{B}, with direction found using the right-hand rule.

Derivation/expression: Consider a charge qq moving with velocity v⃗\vec{v} in a region of uniform magnetic field B⃗\vec{B}. Experimentally, the magnetic force on the charge is found to be proportional to qq, to vv, to BB, and to sin⁡θ\sin\theta (where θ\theta is the angle between v⃗\vec{v} and B⃗\vec{B}), and it is always perpendicular to the plane containing v⃗\vec{v} and B⃗\vec{B}. This is compactly written as the vector (cross) product:

F⃗=q(v⃗×B⃗)\vec{F} = q(\vec{v}\times\vec{B})

Its magnitude is F=qvBsin⁡θF = qvB\sin\theta. Notice that if v⃗\vec{v} is parallel to B⃗\vec{B} (θ=0\theta=0), the force is zero; the force is maximum (F=qvBF=qvB) when v⃗⊥B⃗\vec{v}\perp\vec{B}.

For a straight current-carrying conductor of length ll carrying current II, the equivalent expression is F⃗=Il⃗×B⃗\vec{F} = I\vec{l}\times\vec{B}.

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.