Question

(a) Write the expression for the force, vector F, acting on a charged particle of charge ‘q’, moving with a velocity vector v in the presence of both electric field vector E and magnetic field vector B. Obtain the condition under which the particle moves undeflected through the fields.

(b) A rectangular loop of size l × b carrying a steady current I is placed in a uniform magnetic field vector B. Prove that the torque vector τ acting on the loop is given by vector τ = vector m × vector B, where vector m is the magnetic moment of the loop

Answer 1

(a) Electric force on particle, vector Fe = vector qE

If a charge particle enter'* perpendicular to both the electric and magnetic fields then it may happen that the electric and magnetic forces cancel each other and so particle will pass undeflected.

(b) Torque on a current carrying loop: Consider a rectangular loop PQRS of length l, breadth b suspended in a uniform magnetic field vector B. The length of loop = PQ = RS =l and breadth = QR =SP = b. Let at any instant the normal to the plane of loop make an angle θ with the direction of magnetic field vector B and I be the current in the loop. We know that a force acts on a current carrying wire placed in a magnetic field. Therefore, each side of the loop will experience a force. The net force and torque acting on the loop will be determined by the forces acting on all sides of the loop. Suppose that the forces on sides PQ, QR, RS and SP are vector F1, vector F2, vector F3 and vector F4 respectively. The sides QR and SP make angle (90° - θ) with the direction of magnetic field. Therefore each of the forces vector F2and vector F4 acting on these sides has same magnitude F' = Blb sin(90° - θ) = Blb cosθ. According to Fleming’* left hand rule the forces vector F2 and vector F4 are equal and opposite but their line of action is same. Therefore these forces cancel each other i.e. the resultant of vector F2 and vector F4 is zero. The sides PQ and RS of current loop are perpendicular to the magnetic field, therefore the magnitude of each of forces vector F1 and vector F3 is F = IlBsin90° = IlB.

According to Fleming’* left hand rule the forces vector F1 and vector F3 acting on sides PQ and RS are equal and opposite, but their lines of action are different; therefore the resultant force of vector F1and vector F3 is zero, but they form a couple called the deflecting couple. When the normal to plane of loop makes an angle q with the direction of magnetic field B, the perpendicular distance between F1 and F3 is b sinθ.

∴ Moment of couple or Torque,

τ = (Magnitude of one force F) ´ perpendicular distance =(BIl) × (bsinθ) = I(lb) Bsinθ

But lb =area of loop = A(say)

∴ Torque, τ = IABsinθ

If the loop contains N-turns, then τ = NIABsinθ

In vector form vector τ = NI vector A x vector B.

The magnetic dipole moment of rectangular current loop = M = NIA

∴ vector τ = vector M x vector B

Direction of torque is perpendicular to direction of area of loop as well as the direction of magnetic field i.e., along I vector A x vector B.