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Step 1: List the Known Parameters
• Side of the square loop,
s = 2.0 \times 10^{-2}\,\mathrm{m}.
• Number of turns per centimetre in the solenoid,
n = 50\,\mathrm{turns/cm} = 50/(1\times10^{-2}\,\mathrm{m}) = 5000\,\mathrm{turns/m}.
• Amplitude of the current,
I_0 = 2.5\,\mathrm{A}.
• Angular frequency,
\omega = 700\,\mathrm{rad\,s}^{-1}.
• Permeability of free space,
\mu_0 = 4\pi \times 10^{-7}\,\mathrm{T\,m\,A}^{-1}.
Step 2: Magnetic Field in a Long Solenoid
For a long solenoid carrying current I(t) , the magnetic field inside is given by:
B(t) = \mu_0 \, n \, I(t).
Step 3: Magnetic Flux Through the Square Loop
The area of the square loop is:
A = s^2 = (2.0 \times 10^{-2} \,\mathrm{m})^2
= 4.0 \times 10^{-4}\,\mathrm{m^2}.
If the current in the solenoid varies as
I(t) = I_0 \sin(\omega t) ,
the magnetic flux through the loop at any time t is:
\Phi(t) = B(t) \times A = (\mu_0\, n\, I(t)) \,A.
Step 4: Faraday’s Law of Induction
According to Faraday’s law, the emf induced in the loop is:
\varepsilon(t) = -\,\frac{d\Phi}{dt}.
Substituting \Phi(t) :
\varepsilon(t)
= -\frac{d}{dt}\bigl(\mu_0\,n\,A\,I(t)\bigr)
= -\mu_0\,n\,A\,\frac{dI(t)}{dt}.
Step 5: Differentiate the Current
Since I(t) = I_0 \sin(\omega t) ,
\frac{dI(t)}{dt} = I_0 \,\omega \cos(\omega t).
Thus,
\varepsilon(t)
= -\,\mu_0\,n\,A\,\bigl(I_0\,\omega \cos(\omega t)\bigr).
Step 6: Amplitude of the Induced emf
The induced emf will have a maximum (amplitude) when \cos(\omega t) = 1 . Therefore,
\varepsilon_{\max}
= \mu_0 \, n \, A \, I_0 \, \omega.
Substituting the numerical values:
\varepsilon_{\max}
= (4\pi \times 10^{-7})
\times (5000)
\times (4.0 \times 10^{-4})
\times (2.5)
\times (700).
Step 7: Simplify the Expression
1. Combine \mu_0 and n :
(4\pi \times 10^{-7}) \times 5000 = 20\pi \times 10^{-4}.
2. Multiply by area A :
20\pi \times 10^{-4} \times (4.0 \times 10^{-4})
= 80\pi \times 10^{-8}
= 8\pi \times 10^{-7}.
3. Multiply by I_0 :
8\pi \times 10^{-7} \times 2.5
= 20\pi \times 10^{-7}
= 2\pi \times 10^{-6}.
4. Multiply by \omega :
2\pi \times 10^{-6} \times 700
= 1400\pi \times 10^{-6}
= 1.4\pi \times 10^{-3}.
Using \pi \approx \frac{22}{7} \approx 3.14 ,
\varepsilon_{\max}
\approx 1.4 \times 3.14 \times 10^{-3}
\approx 4.396 \times 10^{-3}\,\mathrm{V}
\approx 4.4 \times 10^{-3}\,\mathrm{V}.
Hence, we can write:
4.4 \times 10^{-3}\,\mathrm{V}
= 44 \times 10^{-4}\,\mathrm{V}.
Step 8: Final Answer
The amplitude of the induced emf is given in the form
x \times 10^{-4}\,\mathrm{V} ,
so
x = 44.
