generic · CBSE Class 12th English Medium · PHYSICS PART-1 · Page 275question

Waves

Chapter 8: Chapter 8 · PHYSICS PART-1 · EN medium

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Now, consider a different surface, which has the same boundary. This is a pot like surface [Fig. . (b)] which nowhere touches the current, but has its bottom between the capacitor plates; its mouth is the circular loop mentioned above. Another such surface is shaped like a tiffin box (without the lid) [Fig. . (c)]. On applying Ampere’s circuital law to such surfaces with the same perimeter, we find that the left hand side of Eq. ( . ) has not changed but the right hand side is zero and not μ0i, since no current passes through the surface of Fig. . (b) and (c).

📖 NCERT Class 12 Physics Part 1 · Page 275

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Now, consider a different surface, which has the same boundary. This is a pot like surface [Fig. . (b)] which nowhere touches the current, but has its bottom between the capacitor plates; its mouth is the circular loop mentioned above.

Another such surface is shaped like a tiffin box (without the lid) [Fig. . (c)]. On applying Ampere’s circuital law to such surfaces with the same perimeter, we find that the left hand side of Eq.

( . ) has not changed but the right hand side is zero and not μ0i, since no current passes through the surface of Fig. . (b) and (c).

So we have a contradiction; calculated one way, there is a magnetic field at a point P; calculated another way, the magnetic field at P is zero. Since the contradiction arises from our use of Ampere’s circuital law, this law must be missing something. The missing term must be such that one gets the same magnetic field at point P, no matter what surface is used. We can actually guess the missing term by looking carefully at Fig.

. (c). Is there anything passing through the surface S between the plates of the capacitor? Yes, of course, the electric field!

If the plates of the capacitor have an area A, and a total charge Q, the magnitude of the electric field E between the plates is (Q/A)/ε0 (see Eq. . ). The field is perpendicular to the surface S of Fig.

. (c). It has the same magnitude over the area A of the capacitor plates, and vanishes outside it. So what is the electric flux ΦE through the surface S ?

Using Gauss’s law, it is

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