A circular insulated copper wire loop is twisted to form two loops of area A and 2 A as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field B → points into the plane of the…

Question

A circular insulated copper wire loop is twisted to form two loops of area $A$ and $2 A$ as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field $\vec{B}$ points into the plane of the paper. At $t = 0$ , the loop starts rotating about the common diameter as axis with a constant angular velocity in the magnetic field. Which of the following options is/are correct?

Aicharya · Accepted Answer

When a conducting loop rotating in a magnetic field $B$ has an angular velocity $ω$ , the flux through the loop of area $A$ at an angle $θ = ω t$ ,

$ϕ = |B| |A| \cos θ$

$= B A \cos (ω t)$

The induced emf,

$ε = - \frac{d ϕ}{d t} = B A ω \sin (ω t)$

So, $ε a n d \frac{d ϕ}{d t} \propto \sin (ω t)$

So, the emf is maximum when, $ω t = θ = \frac{π}{2} .$

Since the emf in the smaller loop will act opposite to that of the larger loop,

$ε_{N e t} = ε_{2 A} - ε_{A} = B (2 A) ω \sin ω t - B (A) ω \sin (ω t)$

$= B (2 A - A) ω \sin ω t = B A ω \sin ω t$

So, the amplitude of the maximum net emf is equal to $B A$ which is same as that in the smaller loop.

A circular insulated copper wire loop is twisted to form two loops of area A and 2 A as shown in the figure.…

Solution