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Homework Statement
There is a uniform but variable magnetic field ##\vec{B}=(B_0 t)(\hat{k})##, in a cylindrical region, whose boundary is described by ##x^2+y^2=a^2##. ##\displaystyle \int_P^{Q} \vec{E} \cdot \vec{dy}## is (see attachment 1)
A)0
B)##\frac{\pi}{4}(B_0 a^2)##
C)##\frac{\pi}{8}(B_0 a^2)##
D)##\frac{\pi}{8}(B_0 a^2)##
Homework Equations
The Attempt at a Solution
(see attachment 2)
The electric field produced will be in anticlockwise direction. At a distance y from P, the electric field can be calculated using Gauss's law.
[tex]E \cdot 2\pi \sqrt{4a^2+y^2}=B_0 \pi a^2[/tex]
[tex]E=\frac{B_0 a^2}{2\sqrt{4a^2+y^2}}[/tex]
Since y is in the vertical direction, we need to only consider the vertical component to evaluate the integral asked in the question.
[tex]\int_{P}^{Q} \vec{E}\cdot \vec{dy}=\int_0^{2a} E\cos \theta dy[/tex]
[tex]=\int_0^{2a} \frac{B_0 a^2}{2\sqrt{4a^2+y^2}} \cdot \frac{2a}{\sqrt{4a^2+y^2}} dy[/tex]
Solving the integral, I get D which is correct.
The problem is I don't understand what the solution booklet has mentioned. The solution goes like this: (see attachment 3)
[tex]\int_{P}^Q \vec{E} \cdot \vec{dy}=\frac{1}{8}[\vec{E} \cdot \vec{dl} \text{ for square } TQRST][/tex]
[tex]=\frac{1}{8} \frac{d[\pi a^4B_0t]}{dt}[/tex]
[tex]=\frac{\pi}{8}B_0 a^2[/tex]
I don't understand how they even got the first two steps.
Any explanation on this would be helpful. Thanks!
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