Show a rectangular loop MNOP being pulled out of a magnetic field with a uniform velocity v by applying an external force F. The length MN is equal to l and the total resistance of the loop is R. Find(a) the current in the loop,(b) the magnetic force on the loop, (c) the external force F needed to move it at constant velocity, (d) the power delivered by the external force and (e) the thermal power developed in the loop.

Explanation

(a) The emf induced in the loop is due to the motion of the wire MN
The emf is `varepsilon= vBl` with the positive end at N and the negative and at M
The current is `I = (varepsilon)(R ) = (vBI)/( R)` in the clockwise direction (b) The magnetic force on the wire MN is `vecF_1 =i vecL X vecB`
The magnitude is `F_1 = ilB = (vB^2l^2)/(R )` and is opposite to the velocity
the forces on the parts of the wire NO and PM, lying in the fijled, cancle each other
The resultant magnetic force on the loop is, therfore, `F_1 = (B^l^2v)/(R )` opposite ot the velocity
(c)To move the loop at a constant velocity, the resultant force on it should be zero
Thus, one should pull the loop with a force ` F =F_1 =(vB^2l^2)/(R )`
(d) The power delivered by the external force is `P =F_V =(v^2B^2l^2)/(R)` (e) The thermal power developed is `P = l ^2R = ((vBI)/(R))^2 R= (v ^2B ^2l^2)/(R)`
We see that the power delivered by the external force is equal to the thermal power developed in the loop
this is consistent with the principle of conservation of energy