NEET-XII-Physics
38: Electromagnetic Induction
- #20-athe contact begins to slide and (b) it has slid through half the length of the rheostat.
FigureAns : For y = L,
`` e=\frac{{\,\mathrm{\,\mu \,}}_{0}N\,\mathrm{\,\pi \,}{a}^{2}a{\mathit{\text{'}}}^{\mathit{2}}\epsilon Rv}{2L({a}^{2}+{x}^{2}{)}^{3/2}(R+r{)}^{2}}`` (b) For y = L/2,
`` \frac{R}{L}y=\frac{R}{2}``
`` \Rightarrow e=\frac{{\mu }_{0}N\pi {a}^{2}a{\text{'}}^{2}}{2L({a}^{2}+{x}^{2}{)}^{3/2}}\frac{\epsilon Rv}{{\left({\displaystyle \frac{R}{2}}+r\right)}^{2}}``
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- #20-bit has slid through half the length of the rheostat.
FigureAns : For y = L/2,
`` \frac{R}{L}y=\frac{R}{2}``
`` \Rightarrow e=\frac{{\mu }_{0}N\pi {a}^{2}a{\text{'}}^{2}}{2L({a}^{2}+{x}^{2}{)}^{3/2}}\frac{\epsilon Rv}{{\left({\displaystyle \frac{R}{2}}+r\right)}^{2}}``
Page No 307: