PHI =BA=\frac{B}{2}\frac{X}{\sqrt{2}}\frac{x}{\sqrt{2}}\] \[\varepsilon =\frac{d\phi }{DT}=\frac{2x}{4}\frac{dx}{dt}B=\frac{x}{2}BV\] \[\varepsilon =\frac{d\phi }{dt}=\frac{2x}{4}\frac{dx}{dt}B=\frac{x}{2}B\frac{d(2y)}{dt}\] \[i=\frac{xBv}{2\lambda (x+\sqrt{2}x)}=\frac{Bv}{\lambda (1+\sqrt{2})}\]

"> PHI =BA=\frac{B}{2}\frac{X}{\sqrt{2}}\frac{x}{\sqrt{2}}\] \[\varepsilon =\frac{d\phi }{DT}=\frac{2x}{4}\frac{dx}{dt}B=\frac{x}{2}BV\] \[\varepsilon =\frac{d\phi }{dt}=\frac{2x}{4}\frac{dx}{dt}B=\frac{x}{2}B\frac{d(2y)}{dt}\] \[i=\frac{xBv}{2\lambda (x+\sqrt{2}x)}=\frac{Bv}{\lambda (1+\sqrt{2})}\]

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There are three wire MO, NO and PQ, wires MO and NO are fixed and perpendicular to each other. Wire PQ moves with a constant velocity v as shown in the figure and resistance per unit length of each wire is\[\lambda \]and magnetic field exists perpendicular and inside the paper then. Which of the following is wrong?

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) 1 year ago

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[d] \[\PHI =BA=\frac{B}{2}\frac{X}{\sqrt{2}}\frac{x}{\sqrt{2}}\] \[\varepsilon =\frac{d\phi }{DT}=\frac{2x}{4}\frac{dx}{dt}B=\frac{x}{2}BV\] \[\varepsilon =\frac{d\phi }{dt}=\frac{2x}{4}\frac{dx}{dt}B=\frac{x}{2}B\frac{d(2y)}{dt}\] \[i=\frac{xBv}{2\lambda (x+\sqrt{2}x)}=\frac{Bv}{\lambda (1+\sqrt{2})}\]

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Posted on 04 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Physics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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