A1000-turn solenoid, is 50 cm long, has a diameter of 25 mm. A 20-turn coil is wound tightly around the center of the solenoid. If the current in the solenoid increases uniformly from 0 to 10A in 1 minute. What will be the induced emf in the short coil during this time?

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rumber of turns ir the souroid, N_{s}=1000urgth of socerid, L_{S}=50 \mathrm{~cm}=0.5 \mathrm{~m}dimefer of sourid, d_{s}=25 \mathrm{~mm}=0.025 \mathrm{~m}rumber of thins of the 1 \mathrm{ii}, \mathrm{N}_{c}=20iritial kurrert, E_{i}=0 AFinal currert, I_{f}=10 \mathrm{~A}cwrge in curreat, \Delta E=F_{f}-E_{i}=10 \mathrm{~A}-O A=10 \mathrm{~A}rime, \Delta t=1 mirer bosThe errf irduced in the wil is gi,ren by,\varepsilon=\frac{\Delta \phi_{B}}{\Delta t}a_{B}=N C B Amono the megretic field dne to the souroid,B=\frac{\mu_{0} N_{s} E}{4}(o, \varepsilon=\frac{\Delta\left(H_{L} B A\right)}{\Delta t}=r_{C} A \frac{\Delta}{\Delta t}\left(\frac{\mu_{0} r_{S} I}{L_{S}}\right)\varepsilon=\frac{\mu_{0} r_{c} N_{3} A}{y}\left(\frac{\Delta \tau}{\Delta f}\right)\begin{array}{l}A=\frac{\pi d_{s}}{4} \\\varepsilon=\frac{\mu_{0} r_{c} r_{s} \cdot \pi d_{s}^{2}}{4 L_{s}}\left(\frac{\Delta \tau}{\Delta t}\right)\end{array}\begin{array}{l}\varepsilon=\frac{4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A} \times 20 \times 1000 \times 0 \times(0.025 \mathrm{~m})^{2}}{4 \times 0.5 \mathrm{~m}} \times \frac{10 \mathrm{~A}}{60 \mathrm{~s}} \\\varepsilon=4.11 \times 10^{-6} \mathrm{~V}\end{array}so, indwed renf 12 the short col; is\varepsilon-4.11 \times 10^{-6} vIn order to find the emf induced in the coil we have to use the formula  ,  \varepsilon=\frac{\Delta \Phi_{B}}{\Delta t}=\frac{\Delta\left(N_{C} B A\right)}{\Delta t} ...