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Given in the question,magnetic field, B=2.0 T.Diameter, d=5.0 \mathrm{~cm}=0.05 \mathrm{~m}Decreasing rate, \frac{d B}{d t}=4.90 \mathrm{~T} / \mathrm{S}.(a) using equation,Electric field inside a solenoid,E=\frac{r}{2} \cdot\left|\frac{d B}{d t}\right|Hare, on the axis r=0, So\begin{aligned}E & =\frac{0}{2} \times 4.9 \\& =0\end{aligned}Thus, Electric field inside the sole noid on the ax is is zepo.(B) Now, for a point at a distance,p=1.40 \mathrm{~cm}=1.40 \times 10^{-2} \mathrm{~m}electric field, E=\frac{r}{2}\left|\frac{d B}{d t}\right|\begin{array}{l}=\frac{1.40 \times 10^{-2}}{2} \times 4.90 \\=3.43 \times 10^{-2} \mathrm{~V} / \mathrm{m}\end{array}Answer:(A) 0 \mathrm{~V} / \mathrm{m}(B) 3.43 \times 10^{-2} \mathrm{~V} / \mathrm{m}Given in the question, magnetic field, B=2.07. Diameter, d=5.0 \mathrm{~cm}=0.05 \mathrm{~m} Decreasing paie, .4901 / \mathrm{s} (a ) using equation Electric field inside a a Solenoid, \mathrm{E} po \mathrm{Z} \mathrm{dB} dt Hare; on the axis \mathrm{P}=0, so \mathrm{E}=\mathrm{Q} \times 4.9= Thus, Electric field inside the solenoid on the axis is zero.(B Now, for a point at a distance, P=1.40 \mathrm{~cm} 1.40 \times 102 \mathrm{~m} electric field, E la at 1.408102 \times 4.90 \mathrm{~dB} P 2 If 2=3.43 \times 102 \mathrm{~V}/ \mathrm{m} ...