# Rate of decrease of the magnetic fields

Let V_B be the rate of decrease of the magnetic fields

$$\frac{dB}{dt}$$

For the 3rd path:
$$\oint E\cdot ds = -\frac{d\phi _B}{dt} = -\frac{{d\phi _B}_1 + {d\phi _B}_2}{dt}$$

$$\phi _B_{(t)} = A_{(t)}B_{(t)}$$
The area is constant, it's only the magnetic field that's changing:
$$\phi _B_{(t)} = \pi R^2(B_0 - V_Bt)$$

Since B1 and B2 are in opposite directions, give one of them a minus sign:
$${\phi _B}_1 + {\phi _B}_2 = \pi R_1^2(B_0 - V_Bt) - \pi R_2^2(B_0 - V_Bt) = \pi (B_0 - V_Bt)(R_1^2 - R_2^2)$$
Take the derivative of that:
$$\frac{{d\phi _B}_1 + {d\phi _B}_2}{dt} = -\pi V_B(R_1^2 - R_2^2)$$

And therefore:
$$\oint E\cdot ds = \pi V_B(R_1^2 - R_2^2)$$

## Answers and Replies

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