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| 184_notes:ind_graphs [2022/12/07 14:37] – valen176 | 184_notes:ind_graphs [2022/12/07 14:43] (current) – valen176 | ||
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| $$V_{ind} = -\frac{d\Phi_b}{dt}$$ | $$V_{ind} = -\frac{d\Phi_b}{dt}$$ | ||
| - | This is saying that the induced current is the **negative slope** of the magnetic flux. In other words, if the magnetic flux is increasing, then $V_{ind}$ will be negative, if the magnetic flux is decreasing, then $V_{ind}$ will be positive, and if the magnetic flux is constant, then $V_{ind} = 0$ | + | This is saying that the induced current is the **negative slope** of the magnetic flux. In other words, if the magnetic flux is increasing, then $V_{ind}$ will be negative, if the magnetic flux is decreasing, then $V_{ind}$ will be positive, and if the magnetic flux is constant, then $V_{ind} = 0$. |
| - | First let's consider when $\Phi_B$ rises and falls linearly with the same magnitude of slope: | + | First let's consider when an example where $\Phi_B$ rises and falls linearly with the same magnitude of slope: |
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| \end{cases} | \end{cases} | ||
| $$ | $$ | ||
| - | Which finally means that $V_{ind}$ | + | Now we can multiply by $-1$ because of the negative sign in Faraday' |
| $$ | $$ | ||
| V_{ind}= | V_{ind}= | ||