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Question: A conducting loop in the shape of a right angled isosceles triangle of height \(10cm\) is kept such ...

A conducting loop in the shape of a right angled isosceles triangle of height 10cm10cm is kept such that the 90{90^ \circ } 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 the counter clockwise direction and increases at a constant rate of 10A/s10A/s Which of the following statements(s) is(are) true?

A. There is a repulsive force between the wire and the loop
B. If the loop is rotated at a constant angular speed about the wire, an additional emf of (μoπ)(\dfrac{{{\mu _o}}}{\pi }) volt is inducted in the wire.
C. The magnitude of induced emf in the wire is (μoπ)(\dfrac{{{\mu _o}}}{\pi }) volt
D. The induced current in the wire is in opposite direction to the current along the hypotenuse

Explanation

Solution

Find the flux through the wire assuming assuming some value of current and rate of change of current is given using that find mutual inductance of the system once you know the mutual inductance use faraday's law to find the magnitude of induced emf and use lenz's law to find the direction of induced emf or induced current.

Complete step by step answer:
Let's take a strip of width dx in the triangle perpendicular to the length of infinite wire


Now,
The flux passing through triangular wire when i current is passing through the infinite wire is
ϕ=B.dA=00.1μoi2πx×2x.dx\phi = \int {B.dA} = \int\limits_0^{0.1} {\dfrac{{{\mu _o}i}}{{2\pi x}}} \times 2x.dx where 0.1 is the height of triangle
ϕ=00.1μoi2πx×2x.dx=[μoiπx]00.1=μoi10π\Rightarrow \phi = \int\limits_0^{0.1} {\dfrac{{{\mu _o}i}}{{2\pi x}}} \times 2x.dx = \left[ {\dfrac{{{\mu _o}i}}{\pi }x} \right]_0^{0.1} = \dfrac{{{\mu _o}i}}{{10\pi }}
We can also write flux as ϕ=Mi\phi = Mi
Therefore, M=μo10πM = \dfrac{{{\mu _o}}}{{10\pi }}
Rate of change of current is given as didt=10A/s\dfrac{{di}}{{dt}} = 10A/s
So induced emf ε=Mdidt\varepsilon = M\dfrac{{di}}{{dt}}
ε=μo10π×10=μoπ\Rightarrow \varepsilon = \dfrac{{{\mu _o}}}{{10\pi }} \times 10 = \dfrac{{{\mu _o}}}{\pi }
If loop rotates then emf is produced which must be opposed and in order to do so force on the loop should be outwards therefore,
There will be repulsion between loop and wire and direction of current will be the same as the direction of current in hypotenuse.

So, the correct answer is “Option A and C”.

Note:
Here finding the value of mutual inductance is crucial and to do that you will have to use integration there is no other shortcut way once you know the value of mutual inductance then simply use faraday's law and lenz's law to get to the answer.The hypotenuse of the triangle is parallel to the wire.