32 Emfs and voltmeters circular wire in Fig. 7.39(a) encircles a solenoid in whi

Question

32 Emfs and voltmeters circular wire in Fig. 7.39(a) encircles a solenoid in which the flux d/dt is increasing at a constant rate 80 out of the page In Fig. 7.39(b) the solenoid has been removed, and a capacitor has been inserted in the loop. The upper plate is positive. The volt- b) age difference between the plates is Co, and this voltage is main- tained by someone physically dragging positive charges from the negative plate to the positive plate (or rather, dragging electrons the other way). So this person is the source of the emf. In Fig. 7.39(c) the above capacitor has been replaced by N little capacitors, each with a voltage difference of Eo/N. The figure shows N = 12, but assume that N is large, essentially infinite. As above, the emf is maintained by people dragging charges from one plate to the other in every capacitor. This setup is similar to the setup in Fig. 7.39(a), in that the emf is evenly distributed around the circuit. By definition, the voltage difference between two points is F given by V, - Va-E ds. This is what a voltmeter mea- sures. For each of the above three setups, find the voltage differ- ence Vb - Va along path 1 (shown in part (a) of the figure), and also the voltage difference Va- Vi along path 2. Comment on the similarities and differences in your results, and also on each of the total voltage drops in a complete round trip.

Explanation / Answer

we have to find Vb - Va for the path 1 and path 2 in the three cases

case 1 : along path 1

Vb - Va = -E*L/ab ( as the points are really close to each other and the potential difference V is uniformly distributed along the length of the wire, here ab is length of the wire ab and L is length of the whoile wire)

along path 2 : as Vb - Va does not depend on the choice of the path used, hence Vb - Va = -EL/ab

Case 2 : Vb - Va in both paths = -E

Case 3 : Vb - Va along both paths = -E/n ( because total potential difference in the whole loop must be zero, so for potential drop of E/n alopng the capacitor, potential drop in the rest of the loop must be -ve of this to make the sum 0)

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