A conducting loop in the shape of right angled isosceles triangle of height...

A conducting loop in the shape of right angled isosceles triangle of height 10 cm is kept such that the $90^{o}$ 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?

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

(\frac{μ_{0}}{π})

volt is induced in the wire.

The magnitude of induced emf in the wire is

(\frac{μ_{0}}{π})

volt

Solution:

From lenz's law it can be understood that due to induced current in wire flux will change in loop thats why it will be repelled by the loop

Also by direction of induced electric field, the current in the wire is in the same direction of current along the hypotenuse.
Flux through triangle if wire has current $i_{1}$

$ϕ_{2} = \int_{0}^{0.1} B d A$

$= \int_{0}^{0.1} (\frac{μ_{0} i_{1}}{2 π x}) (2 x d x) = \frac{μ_{0} i_{1}}{10 π}$

$\Rightarrow ϕ_{2} = M i_{1}$
$\Rightarrow$ Mutual inductance $M = \frac{μ_{0}}{10 π}$
Induced emf in wire

$ε_{1} = M \frac{d i_{2}}{d t}$

$= \frac{μ_{0}}{10 π} \frac{d i_{2}}{d t} = \frac{μ_{0}}{10 π} \times 10 = \frac{μ_{0}}{π}$

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A conducting loop in the shape of right angled isosceles triangle of height...

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A conducting loop in the shape of right angled isosceles triangle of height...

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