(a) Two straight long parallel conductors carry currents I1 and I2 in the same direction. Deduce the expression for the force per unit length between them and depict the pattern of magnetic field lines around them. (b) A rectangular current-carrying loop EFGH is kept in a uniform magnetic field. (i) What is the direction of the magnetic moment of the current loop? (ii) When is the torque acting on the loop (A) maximum and (B) zero?
Solution
(a) Field of wire 1 at wire 2 is B1=2πdμ0I1, giving force per unit length ℓF=2πdμ0I1I2 (attractive for same-direction currents); the field lines between the wires oppose and cancel, drawing them together. (b)(i) The magnetic moment m=NIA is normal to the plane of the loop, sense given by the right-hand rule. (ii) τ=mBsinθ: it is \textbf{maximum} when θ=90∘ (plane of loop parallel to B, i.e. m⊥B) and \textbf{zero} when θ=0∘ (m∥B, plane perpendicular to B).
CBSE Class 12 Board · 2009 · All India · Subjective
