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  4. A circuit ABCD is held perpendicular to the uniform magnetic field of B=5× 10- 2T extending over the region PQRS and directed into the plane of the paper. The circuit is moving out of the field at a uniform speed of 0.2ms- 1 for 1.5s . During this time, the current in the 5Ω resistor is <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-lem6xcmkpwmcoz9s.jpg />

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Q. A circuit $ABCD$ is held perpendicular to the uniform magnetic field of $B=5\times 10^{- 2}T$ extending over the region $PQRS$ and directed into the plane of the paper. The circuit is moving out of the field at a uniform speed of $0.2ms^{- 1}$ for $1.5s$ . During this time, the current in the $5\Omega$ resistor is
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NTA AbhyasNTA Abhyas 2022

Solution:

$I=\frac{B \ell v}{R}$
$I=\frac{5 \times 10^{- 2} \times 0 .3 \times 0 .2}{5}A=0.6mA.$
Area and flux are decreasing. So, current flows to increase the flux. Clearly, current should be clockwise. So, it flows from $B$ to $C$ through $5\Omega$ .