A) \[6\times {{10}^{-4}}\,Volt\]
B) \[4.8\times {{10}^{-2}}Volt\]
C) \[6\times {{10}^{-2}}\,Volt\]
D) 48 kV
Correct Answer: A
Solution :
Magnetic field of solenoid, \[{{B}_{1}}\,\,=\,\,\frac{{{\mu }_{0}}\,{{N}_{1}}\,{{i}_{1}}}{\ell }\] Magnet flux of coil, \[{{\phi }^{2}}\,=\,{{N}_{2}}\,{{B}_{1}}\,{{A}_{2}}\,=\,{{N}_{2}}\,\left( \frac{{{\mu }_{0}}\,{{N}_{1}}\,{{i}_{1}}}{\ell } \right)\,{{A}_{2}}\] \[As\,\,{{\phi }^{2}}\,=\,\,M\,{{i}_{1}}\,so\,\,M\,\,=\,\,\frac{{{\phi }_{2}}}{{{i}_{1}}}\,\,=\frac{{{\mu }_{0}}\,{{N}_{1}}\,{{N}_{2\,}}{{A}_{2}}}{\ell }\,\] \[\therefore \] induced emf, \[\left| e \right|\,\,\,=\,\,\,\frac{{{\mu }_{0}}\,{{N}_{1}}\,{{N}_{2}}\,{{A}_{2}}}{\ell }\,\,\times \,\,\frac{d{{i}_{1}}}{dt}\] \[=\,\,\frac{4\pi \times {{10}^{-7}}\times 2000\times 300\times 1.2\times {{10}^{-3}}}{0.30}\,\times \frac{4}{0.25}\] \[=\text{ }4.8\,\,\times \,\,{{10}^{-}}^{2}\,Volt\]
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