PL explain...with directions ?PL don't copy as I did not understand ?torque n force find ..I can't find the second force

PL explain...with directions ?PL don't copy as I did not understand ?torque n force find ..I can't find the second force goN1a2 2 [a2 + (3 a) 213/2 20 (10016) = 10.Jfö - 100 -0-0 It is obvious that all the values are in good agreement with the theore Write the expression for the magnetic moment (m) due to a planar square loop of side carrying a steady current I in a vector form. In the given figur this loop is placed in a horizontal plane near a long straight conductor carrying a steady current 11 at a distance I as shown. Give reasons to explain that the loop will experience a net force but no torque. W rite the expression for this force acting op the loop. ICBSE Delhi 20101 Magnetic moment due to a planar square loop of side I carrying current I is m = IA 2 For square loop A = I m = 1 12 h where h is unit vector-normal tonoop. Magnetic field due to current carrying wire at the location of loop is directed downward perpendicular to plane of loop. Force on QR and SP are equal and opposite, so net force on these sides is zero. Force on side P Q, F PQ 110/1 21t1 2 It From on side RS FRS = go/ (4) 2rt (21) Net force F FPQ x FRS = 47t Torque = r x F = zero 4Tt s lb0_, 3. A parallel plate capacitor is being charged by a time varying current. Explain briefly how Ampere' circuital law is generalized to incorporate the effect due to the displacement current. (A1) 2011 Ans. Consider a parallel plate capacitor, being charged by a battery. A time varying current flowing throug the capacitor. If we consider only the conduction current 1, then we apply Ampere's Circuital L.aw to closed loops Cl and C2, then we get amideå

Dear Student
The magnetic moment  , where, is the unit vector along the normal to the surface of the loop.
The attractive force per unit length of the loop is that is on side PQ

Force is attractive as direction of current is same as direction of current in wire

The repulsive force per unit length of the loop is that is on side RS

​​​​​​​Force is repulsive as direction of current is opposite as direction of current in wire
On other two sides (PS and QR) of loop force will get cancel out as force will be equal and opposite in direction

Net force = Fnet= Fa-Fr

 
The net force acting on the loop is not zero, because the attractive force is greater than the repulsive force.
The torque on the loop is given by,
, where, A is the area vector of the loop and   is the angle between the area vector and the magnetic field.
Here, the area vector is parallel to the magnetic field, so  =00. Therefore,
The torque acting on the loop is zero.

Regards

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