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(a) Write the expression for the force, vecF, acting on a charged particle of charge 'q', moving with a velocity vecV in the presence of both electric field vecE and magnetic field vecB. Obtain the condition under which the particle moves undeflected through the fields. (b) A rectangular loop of size lxxb carrying a steady current I is placed in a uniform magnetic field vecB. Prove that the torque vectau acting on the loop is given by vectau=vecmxxvecB," where "vecm is the magnetic moment of the loop. |
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Answer» Solution :(a) The direction of `VECE,vecv and vecB` are such that `vecE bot vecv, vecE bot vecB and vec v` is not PARALLEL or anti-parallel to `vecB.` Force on charge due to electric field `vecE, vecF = q vecE.` Force on charge due to magnetic field `vecB, vecF = q vecv xx vecB` If PARTICLE moves undeflected then magnitude of both the forces are equal i.e. `|q, vec E|= |q vecVxx vecB |` `rArr" "qE=qvB sin 90^(@)""rArr""qE-qvBrArrv=(B)/(E)` (b) `vectau = NI vecAxxvec A= vecMxxvecB` `"where "vecM=NIvecA` is magnetic moment of LOOP. The unit of magnetic moment in S.I. system is AMPERE `xx"metre"^(2)(Am^(2))`. 
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