- #1
Oblakastouf
- 18
- 0
Let V_B be the rate of decrease of the magnetic fields
[tex]\frac{dB}{dt}[/tex]
For the 3rd path:
[tex]\oint E\cdot ds = -\frac{d\phi _B}{dt} = -\frac{{d\phi _B}_1 + {d\phi _B}_2}{dt}[/tex]
[tex]\phi _B_{(t)} = A_{(t)}B_{(t)}[/tex]
The area is constant, it's only the magnetic field that's changing:
[tex]\phi _B_{(t)} = \pi R^2(B_0 - V_Bt)[/tex]
Since B1 and B2 are in opposite directions, give one of them a minus sign:
[tex]{\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)[/tex]
Take the derivative of that:
[tex]\frac{{d\phi _B}_1 + {d\phi _B}_2}{dt} = -\pi V_B(R_1^2 - R_2^2)[/tex]
And therefore:
[tex]\oint E\cdot ds = \pi V_B(R_1^2 - R_2^2)[/tex]
[tex]\frac{dB}{dt}[/tex]
For the 3rd path:
[tex]\oint E\cdot ds = -\frac{d\phi _B}{dt} = -\frac{{d\phi _B}_1 + {d\phi _B}_2}{dt}[/tex]
[tex]\phi _B_{(t)} = A_{(t)}B_{(t)}[/tex]
The area is constant, it's only the magnetic field that's changing:
[tex]\phi _B_{(t)} = \pi R^2(B_0 - V_Bt)[/tex]
Since B1 and B2 are in opposite directions, give one of them a minus sign:
[tex]{\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)[/tex]
Take the derivative of that:
[tex]\frac{{d\phi _B}_1 + {d\phi _B}_2}{dt} = -\pi V_B(R_1^2 - R_2^2)[/tex]
And therefore:
[tex]\oint E\cdot ds = \pi V_B(R_1^2 - R_2^2)[/tex]