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Question Number 42422 by ajfour last updated on 25/Aug/18

Commented by ajfour last updated on 25/Aug/18

Q.42278  solution  (rishav′s question)

$${Q}.\mathrm{42278}\:\:{solution}\:\:\left({rishav}'{s}\:{question}\right) \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 25/Aug/18

pls explain...

$${pls}\:{explain}... \\ $$

Answered by ajfour last updated on 25/Aug/18

considering area A_(OAB) ^�   :  (dφ_B /dt)= (l/2)(dB/dt)(√(R^2 −(l^2 /4))) = ∮_(OAB) E^� .ds^�   ⇒     (l/2)(√(R^2 −(l^2 /4))) = ∫_O ^(  A) E^� .ds^�                                    +∫_A ^(  B) E^� .ds^�  +∫_B ^(  O) E^� .ds^�   ⇒ (l/2)(dB/dt)(√(R^2 −(l^2 /4))) = 0+(V_A −V_B )+0  hence induced emf along rod ends      V_A −V_B  = (1/2)(dB/dt)(√(R^2 −(l^2 /4))) .  Alternate method :  Applying Faraday′s law around  circle of radius r.  ∮E^� .ds^�  = (dφ_B /dt) = πr^2 (dB/dt)  E(2πr)=πr^2 (dB/dt)  E = (r/2)(dB/dt)  let ⊥ distance of rod from centre  be h .  Then along the rod  V_A +∫_A ^(  B) E^� .ds^�  =V_B   ⇒ V_B −V_A =(1/2)(dB/dt)∫_A ^(  B)  r(ds)cos θ      V_(BA) =(1/2)(dB/dt)∫_(−α) ^(  α) (hsec θ)ds(cos θ)              =(1/2)(dB/dt)×h∫_A ^(  B) ds             =(1/2)(dB/dt)×hl             =(l/2)(dB/dt)(√(R^2 −(l^2 /4))) .

$${considering}\:{area}\:\bar {{A}}_{{OAB}} \:\:: \\ $$$$\frac{{d}\phi_{{B}} }{{dt}}=\:\frac{{l}}{\mathrm{2}}\frac{{dB}}{{dt}}\sqrt{{R}^{\mathrm{2}} −\frac{{l}^{\mathrm{2}} }{\mathrm{4}}}\:=\:\oint_{{OAB}} \bar {{E}}.{d}\bar {{s}} \\ $$$$\Rightarrow\:\:\:\:\:\frac{{l}}{\mathrm{2}}\sqrt{{R}^{\mathrm{2}} −\frac{{l}^{\mathrm{2}} }{\mathrm{4}}}\:=\:\int_{{O}} ^{\:\:{A}} \bar {{E}}.{d}\bar {{s}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\int_{{A}} ^{\:\:{B}} \bar {{E}}.{d}\bar {{s}}\:+\int_{{B}} ^{\:\:{O}} \bar {{E}}.{d}\bar {{s}} \\ $$$$\Rightarrow\:\frac{{l}}{\mathrm{2}}\frac{{dB}}{{dt}}\sqrt{{R}^{\mathrm{2}} −\frac{{l}^{\mathrm{2}} }{\mathrm{4}}}\:=\:\mathrm{0}+\left({V}_{{A}} −{V}_{{B}} \right)+\mathrm{0} \\ $$$${hence}\:{induced}\:{emf}\:{along}\:{rod}\:{ends} \\ $$$$\:\:\:\:{V}_{{A}} −{V}_{{B}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\frac{{dB}}{{dt}}\sqrt{{R}^{\mathrm{2}} −\frac{{l}^{\mathrm{2}} }{\mathrm{4}}}\:. \\ $$$${Alternate}\:{method}\:: \\ $$$${Applying}\:{Faraday}'{s}\:{law}\:{around} \\ $$$${circle}\:{of}\:{radius}\:\boldsymbol{{r}}. \\ $$$$\oint\bar {{E}}.{d}\bar {{s}}\:=\:\frac{{d}\phi_{{B}} }{{dt}}\:=\:\pi{r}^{\mathrm{2}} \frac{{dB}}{{dt}} \\ $$$${E}\left(\mathrm{2}\pi{r}\right)=\pi{r}^{\mathrm{2}} \frac{{dB}}{{dt}} \\ $$$${E}\:=\:\frac{{r}}{\mathrm{2}}\frac{{dB}}{{dt}} \\ $$$${let}\:\bot\:{distance}\:{of}\:{rod}\:{from}\:{centre} \\ $$$${be}\:{h}\:. \\ $$$${Then}\:{along}\:{the}\:{rod} \\ $$$${V}_{{A}} +\int_{{A}} ^{\:\:{B}} \bar {{E}}.{d}\bar {{s}}\:={V}_{{B}} \\ $$$$\Rightarrow\:{V}_{{B}} −{V}_{{A}} =\frac{\mathrm{1}}{\mathrm{2}}\frac{{dB}}{{dt}}\int_{{A}} ^{\:\:{B}} \:{r}\left({ds}\right)\mathrm{cos}\:\theta \\ $$$$\:\:\:\:{V}_{{BA}} =\frac{\mathrm{1}}{\mathrm{2}}\frac{{dB}}{{dt}}\int_{−\alpha} ^{\:\:\alpha} \left({h}\mathrm{sec}\:\theta\right){ds}\left(\mathrm{cos}\:\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\frac{{dB}}{{dt}}×{h}\int_{{A}} ^{\:\:{B}} {ds} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\frac{{dB}}{{dt}}×{hl} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{{l}}{\mathrm{2}}\frac{{dB}}{{dt}}\sqrt{{R}^{\mathrm{2}} −\frac{{l}^{\mathrm{2}} }{\mathrm{4}}}\:. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 25/Aug/18

excellent...bah darun..

$${excellent}...{bah}\:{darun}.. \\ $$

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