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Derive an expression for torque on a current loop placed in a magnetic field . |
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Answer» Solution :Consider a single rectangular loop PQRS kept in a uniform magnetic field`vecB` . Leta and b bethe LENGTH and breadth of the rectangular loop respectively . LET` hatn` be the unit VECTOR normal to the plane of the current loop . This unit vector `hat n`completely describes the orientation of the loop. Let `vecB`directed from north pole to south pole of the magnet as shown in Figure.  When an electric current is sent through the loop , the net force acting is zero but there will be net torque acting on it. For the sake of UNDERSTANDING, we shall consider two configurations of the loop , (i) unit vector `hatn`points perpendicular to the field (ii) unit vector points at an angle `theta ` with the field. when unit vector `hat n` is perpendicular to the field In the simple configuration , the unit vector `hatn` is perpendicular to the field and plane of the loop is lying on xy plane as shown in Figure. Let the loop be divided into four sections PQ, QR , RS and SP. The Lorentz force on each loop can be calculated as follows : (a) Force on section PQ , `veci = - a hatj " and " vecB = B hati ` ` vecF_("PQ")= vec(Il) xx vecB ( hatj xx hati) = I a Bhatk ` Since the unit vector normal to the plane ` hat n` is along the direction of ` hatk ` (b) The force on section QR `vecl = vec(bi) " and " vecB = B hat i ` ` vecF_("QR") =vec(Il) xx vecB = - I bB ( hati xx hati) = vec0 ` ( c )The force on section RS `vecl = a hatj " and " vecB = B hati ` ` vecF_("RS") = vec(Il) xx vecB = I aB ( hatj xx hati) = - I a B hatk ` Since, theunit vector normal to the plane is along the direction of ` - hat k `. (d) The force on section SP `vecl = - b hatj " and " vecB = B hati ` ` vecF_("SP") = vec(Il)xx vecB = - IbB ( hati xx hati) = vec0 ` The net forceon the rectangular loop is `vecF_("net") = vecF_("PQ") + vecF_("FS") + vecF_("SP") ` ` vecF_("net") = I a B hatk + vec0 - IaB hatk + vecF_("net") = vec0 ` Hence, the net force on the rectangular loop in this configuration is zero . Now let us calculate the net torque due to these dorces about an AXIS passing through the center `vectau_("net") = sum_(i=1)^(4) vectau_(i) = sum_(i=1)^(4) vecr_(i) xx vecF_(i) ` ` = (b/2 IaB + 0 + b/2 Ia B + 0 ) hatj ` ` vectau_("net") = abIB hatj` Since, A = ab is the area of the rectangular loop PQRS, therefore, the net torque for this configuration is ` vectau_("net") = ABI hatj` |
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