Test Index

JEE Advanced 2016 Paper 1

Section: Physics

Question : 7 of 54

Marks: +1, -0

A conducting loop in the shape of 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 electrically insulated from the loop. The hypotenuse of the triangle is parallel to the wire. The current in the triangular loop is in counterclockwise direction and increased at constant rate of 10 A

s^{-1}

.Which of the following statement(s) is (are) true?

[JEE Adv 2016 P1]

The induced current in the wire is in opposite direction to the current along the hypotenuse.
There is a repulsive force between the wire and the loop.
If the loop is rotated at a constant angular speed about the wire, an additional emf of $\left(\frac{\mu_0}{\pi}\right)$ volt is induced in the wire.
The magnitude of induced emf in the wire is $\left(\frac{\mu_0}{\pi}\right)$ volt.

Validate

Solution:

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 = \int\limits_{0}^{0.1} \left(\frac{\mu_0 i}{2\pi x}\right) = (2xdx) = \frac{\mu_0 i}{10\pi}

⇒Mutual inductance=

\frac{\mu_0}{10\pi}

Induced emf in wire=

\frac{\mu_0}{10\pi}\frac{di}{dt} = \frac{\mu_0}{10\pi} \times 10 = \frac{\mu_0}{\pi}

Go to Question:

Prev Question

Next Question