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PhysicsEMI/ACNon-conservative electric fieldHard2 minPYQ_2020

PhysicsHardsingle choice

The circuit shown in the figure consists of two resistances $R_{1}$ & $R_{2}$ connected to two ideal voltmetres $V_{1}$ & $V_{2}$ . Assume that a voltmeter reads $∆ V = - \int_{a}^{b} \vec{E} . d \vec{l}$ between its terminals. A time-varying magnetic field $B (t) = B_{0} t$ (where $B_{0}$ is a positive constant of proper dimensions and $t$ is time) exists in a circular region of radius a and it is directed into the plane of the figure. The reading of voltmeter $V_{2}$ is

Options:

Answer:

Solution:

$ϕ = π a^{2} B_{0} t$

$ε = - \frac{d ϕ}{d t}$

$\Rightarrow ε = π a^{2} B_{0}$ (indeed emf)

$i = \frac{π a^{2} B_{0}}{R_{1} + R_{2}}$ (induced current)

$∆ V_{1} = - \int_{a}^{b} \vec{E} . d \vec{l} = - (- i R_{1}) = \frac{π a^{2} B_{0} R_{1}}{R_{1} + R_{2}}$

$∆ V_{2} = - \int_{a}^{b} \vec{E} . d \vec{l} = - (- i R_{2}) = - \frac{π a^{2} B_{0} R_{2}}{R_{1} + R_{2}}$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Non-conservative electric field

⏱ 2mℹ️ Source: PYQ_2020

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