Question

A conducting loop in the shape of a right angled isosceles triangle of height $10$ cm is kept such that the ${90}^{\circ }$ 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 counter clockwise direction and increased at a constant rate of $10{As}^{-1}$. Which of the following statement (s) is (are) true?

• A. There is a repulsive force between the wire and the loop
• B. The magnitude of induced emf in the wire is $\left(\frac{{\mu }_0}{\pi }\right)$ volt
• C. 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
• D. The induced current in the wire is in opposite direction to the current along the hypotenuse

Answer 1

Answer: (A), (B)

The magnitude of induced emf in the wire is $\left(\frac{{\mu }_0}{\pi }\right)$ volt


As given, $\frac{di}{dt}=10A/s$
Now finding the flux passing through the infinite long (AB) wire.
so finding the flux passing through its small area $dx$.
$d\varphi =\frac{{\mu }_{0}i}{2\pi x}\times 2xdx$ ....(1)
by integrating above equation.
$\int d\varphi ={\int }_0^{0.1}\frac{{\mu }_{0}i}{\pi x}dx$
$\varphi =\frac{{\mu }_{0}i}{\pi }[x{]}_0^{0.1}=\frac{{\mu }_{0}i}{10\pi }$
$\therefore M=\frac{{\mu }_{0}i}{10\pi }$
As given, $\frac{di}{dt}=10A/s$
$e=M\frac{di}{dt}\Rightarrow e=\frac{di}{dt}\times 10$
$e=\frac{{\mu }_0}{\pi }$

Hence option (B) matches correctly and because conductor is not rotating so option (C) is wrong.
In the given loop dirction of current in the wire and in the hypotenuse are same, hence option (D) is wrong.
As the current in the triangular wire is decreasing the induced current in AB is in the same direction as the current in the hypotenuse of the triangular wire. Therefore force will be repulsive. So option (D) is correct.