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The flux linked with a coil at any insta...

The flux linked with a coil at any instant 't' is given by `phi = 10t^(2) - 50t +250`
The induced emf at `t = 3s` is

A

10 V

B

190 V

C

`-190 V`

D

`-10 V`

Text Solution

AI Generated Solution

The correct Answer is:
To find the induced electromotive force (emf) at \( t = 3 \) seconds, we will follow these steps: ### Step 1: Write down the expression for magnetic flux The magnetic flux \( \Phi \) linked with the coil is given by: \[ \Phi(t) = 10t^2 - 50t + 250 \] ### Step 2: Differentiate the flux with respect to time The induced emf (\( E_{\text{induced}} \)) is given by Faraday's law of electromagnetic induction: \[ E_{\text{induced}} = -\frac{d\Phi}{dt} \] Now we need to differentiate \( \Phi(t) \): \[ \frac{d\Phi}{dt} = \frac{d}{dt}(10t^2 - 50t + 250) \] Using the power rule for differentiation: \[ \frac{d\Phi}{dt} = 20t - 50 \] ### Step 3: Substitute \( t = 3 \) seconds into the derivative Now we will substitute \( t = 3 \) seconds into the expression we found for \( \frac{d\Phi}{dt} \): \[ \frac{d\Phi}{dt} \bigg|_{t=3} = 20(3) - 50 \] Calculating this gives: \[ \frac{d\Phi}{dt} \bigg|_{t=3} = 60 - 50 = 10 \] ### Step 4: Calculate the induced emf Now we can find the induced emf: \[ E_{\text{induced}} = -\frac{d\Phi}{dt} = -10 \] ### Final Answer Thus, the induced emf at \( t = 3 \) seconds is: \[ E_{\text{induced}} = -10 \text{ volts} \] ---
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