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(a) Write the expression for the force vec(F)acting on a particleof mass m and charge q moving with velocityvecVin a magneticfield vecB .Under whatconditionswill it move in (i) a circular path and (ii) a helical path ? (b) Show that the kinetic energy of the particle moving a magneticfield remains constant. |
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Answer» Solution :(a) Expression for magneto Lorentz Force is `vec(F)= q(vecv xx VECB)`, where `vecF`is forceexperienced by charge q movingwith velocity `vecV`in magneticfield `vec(B)`. If `vec(B)` is perpendicular to `vec(V)` then `theta = 90^(@)`and from `vecF= q(vecv xx vecB) = qVsin theta = qvB` This force is always perpendicular to the direction of motion of particle as well asmagnetic well. Thus, the path of chargedparticle is circular. When the component of velocity of the chargedparticle is parallel to the direction of force of `vec(E)` then `F= qvB sin theta = 0`and particle moves in a straightpath force experiencedby perpendiculareffect of bothcomponentstake theparticle in helical path. (b) From the relation, `(mv^(2))/(r) = qvB rArr r = (mv)/(qB)` and `T = (2pir)/(v)= (2pi)/(v) xx (mv)/(qB) = (2pim)/(qB) rArr T = (2nm)/(qB)` So, `v = (qB)/(2pir)`i.e, periodic frequency is independentof speed of particle and RADIUS of circular path `omega =2piv= (qB)/(m)` So, speed hence KE of particlein circularpath remains same. Also from the workenergy theorem change in KE is equalto WORK DONE and work done in this, case is zero. |
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