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Q.Force acting on a charged particle when it moves in a combined electric and magnetic field is known as Lorentz force.

(a) A charged particle is released from rest in a region of steady and uniform electric and magnetic fields, which are parallel to each other. What will be the nature of the path followed by the charged particle? Explain your answer.
(b) A rectangular loop carrying a steady current is placed in a uniform magnetic field. Obtain the expression for the torque acting on the loop. (Scores : 2+3)
Kerala DhseKerala DHSE Plus Two Board 2013Subjective· 5mImportance★★★★★
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With E parallel to B and the particle starting from rest, the magnetic force stays zero throughout (since v is always along B), so the particle moves in a straight line, uniformly accelerated by E alone; a current loop in a uniform B, however, experiences a torque τ = NIAB sinθ = m×B from the couple formed by the forces on its opposite current-carrying sides.

  1. Path of a charged particle released from rest in parallel E and B At the instant of release, v=0v = 0, so the magnetic force F⃗B=qv⃗×B⃗=0\vec F_B = q\vec v \times \vec B = 0; only the electric force qE⃗q\vec E acts, accelerating the particle along the direction of E⃗\vec E (which is also the direction of B⃗\vec B, since they are parallel). As the particle speeds up, its velocity v⃗\vec v remains directed along this same line (it only ever has a component along E⃗\vec E, since nothing pushes it sideways). But v⃗\vec v is then always parallel (or antiparallel) to B⃗\vec B, so v⃗×B⃗=0\vec v \times \vec B = 0 at every instant — the magnetic force never acts, however fast the particle moves. So the particle experiences only the constant electric force throughout, and moves in a straight line along the common direction of E⃗\vec E and B⃗\vec B, with uniform acceleration a=qE/ma = qE/m.
  2. Torque on a rectangular current loop in a uniform magnetic field Consider a rectangular loop PQRS of sides aa (= QR = SP) and bb (= PQ = RS), carrying current II, free to rotate about a vertical axis through the midpoints of QR and SP, placed in a uniform field B⃗\vec B such that the normal to the loop makes angle θ\theta with B⃗\vec B.
  • The sides QR and SP (each of length aa, parallel to the rotation axis and to B⃗\vec B's "vertical" component in this arrangement) experience forces that act along the axis and produce no torque about it (they may stretch/compress the loop but don't rotate it, and are often equal & opposite along the same line).
  • The sides PQ and RS (each of length bb) carry current II perpendicular to B⃗\vec B's relevant component, so each experiences a force of magnitude:

F=BIbF = BIb

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