1. Given data

• Resistance of the closed circuit:
  $R = 20\,\Omega$
• Magnetic flux as a function of time:
  $\phi(t) = 8t^2 - 9t + 5 \quad \text{(weber)}$
• Time at which current is required:
  $t = 0.25\,\text{s}$

2. Use Faraday’s law

The induced emf is $e = -\frac{d\phi}{dt}$ So its magnitude is $|e| = \left|\frac{d\phi}{dt}\right|$

Differentiate the flux function: $\frac{d\phi}{dt} = \frac{d}{dt}(8t^2 - 9t + 5) = 16t - 9$

3. Find emf at $t=0.25\,\text{s}$

$\frac{d\phi}{dt}\Big|_{t=0.25} = 16(0.25) - 9 = 4 - 9 = -5$

Hence magnitude of induced emf is $|e| = |-5| = 5\,\text{V}$

4. Use Ohm’s law to find current

$I = \frac{|e|}{R} = \frac{5}{20} = 0.25\,\text{A}$

5. Convert to mA

$0.25\,\text{A} = 250\,\text{mA}$

Final Answer

$\boxed{250}$

The derived answer matches the stored correct answer.