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Question : (a) Calculate the inductance of an air-core solenoid contain...

(a) Calculate the inductance of an air-core solenoid containing 300 turns if the length of the solenoid is 25.0 \mathrm{~cm} and its cross-sectional area is 100 \mathrm{~cm}^{2}

 

 

 

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Solution Using Equation 32.4, we obtain

L =\frac{\mu_{0} N^{2} A}{\ell} =\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right) \frac{(300)^{2}\left(4.00 \times 10^{-4} \mathrm{~m}^{2}\right)}{25.0 \times 10^{-2} \mathrm{~m}} =1.81 \times 10^{-4} \mathrm{~T} \cdot \mathrm{m}^{2} / \mathrm{A}=0.181 \mathrm{mH}

(b) Calculate the self-induced emf in the solenoid if the current through it is decreasing at the rate of 50.0 \mathrm{~A} / \mathrm{s}.

Using Equation 32.1 and given that d / d t=-50.0 \mathrm{~A} / \mathrm{s}, we obtain

\varepsilon_{L} =-L \frac{d I}{d t}=-\left(1.81 \times 10^{-4} \mathrm{H}\right)(-50.0 \mathrm{~A} / \mathrm{s}) =9.05 \mathrm{mV}