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Q.

In figure a wire perpendicular to a long straight wire is moving parallel to the later with a speed  υ=10​  m/s in the direction of the current flowing in the later. The current is 10A , what is the magnitude of the potential difference between the ends of the moving wire is  2×10x ln (y)V then  x+y5=?

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Detailed Solution

B=μ02πirdV=Bvdr=μ0iv2πdrrV=r1r2dV=μ0iv2πInr2r1

V=4π×10-7(10)(10)2πln0.01+0.090.01=2×10-5ln(10)V

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In figure a wire perpendicular to a long straight wire is moving parallel to the later with a speed  υ=10​  m/s in the direction of the current flowing in the later. The current is 10A , what is the magnitude of the potential difference between the ends of the moving wire is  2×10−x ln (y)V then  x+y5=?