A conducting loop which is in the shape of a right-angled isosceles triangle of height 10 cm is kept such that the 90 ° vertex is very close to an infinitely long conducting wire (see the figure). The wire is

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Correct answer: D. (μ0)/(π)

A conducting loop which is in the shape of a right-angled isosceles triangle of height 10 cm is kept such that the 90 °

by direction of induced electric field, current in wire is in same direction of current along the hypotenuse.

Flux through triangle if wire have current i

= \mathop{∫}_{0}^{0.1}\left(\dfrac{{\mu}_{0}i}{2\pix}\right)\left(2xdx\right) = \dfrac{{\mu}_{0}i}{10\pi}

⇒

Mutual inductance

= \dfrac{{\mu}_{0}}{10\pi}

Induced emf in wire

= \dfrac{{\mu}_{0}}{10\pi}\dfrac{di}{dt} = \dfrac{{\mu}_{0}}{10\pi} \times 10 = \dfrac{{\mu}_{0}}{\pi}

A conducting loop which is in the shape of a right-angled isosceles...

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