Question

A right-angled triangle abc, made from a metallic wire, moves at a uniform speed v in its plane as shown in the figure. A uniform magnetic field B exists in the perpendicular direction. Find the emf induced:

(a) in the loop abc,

(b) in the segment bc,

(c) in the segment ac, and

(d) in the segment ab.

Answer 1

Step 1: Given data

1. The velocity of the coil is $V.$
2. The magnitude of the magnetic field is $B.$

Step 2: Formula Used

1. The electromotive force is proportional to the change in magnetic flux with respect to time.
2. If a conductor of length L is moving in a magnetic B, then the induced emf in it is defined by the form, $\overrightarrow{E}=B\overrightarrow{V}\times \overrightarrow{L}$.
3. According to vector algebra, the cross product of $\overrightarrow{V}\times \overrightarrow{L}$ is zero when V and L are parallel and $VL$ when both are perpendicular.

Step 3 : Calculation

1. As we know, the electromotive force induced on a segment is $\overrightarrow{E}=B\overrightarrow{V}\times \overrightarrow{L}$.
2. In the case of bc, the velocity and length of bc are perpendicular to each other. So, the cross-product of V and L is $VL$.
3. The emf induced in the segment bc is $E_{bc}=BV\left(bc\right)$.

Therefore, the emf induced in the segment bc is $E_{bc}=BV\left(bc\right)$.

(1) EMF induced on the segment ac

1. We know, the electromotive force induced on a segment is $\overrightarrow{E}=B\overrightarrow{V}\times \overrightarrow{L}$.
2. In this case, the velocity of the coil and length of ac is parallel to each other. So, the cross-product of V and L is zero.

Therefore, the emf induced in the segment ac is zero.

(2) EMF induced on the segment ab

1. The induced emf in the segment ab is $E_{ab}=-BV\left(bc\right)$ because its component along cb is perpendicular to the velocity, which will exist and another component of ab along ac will be zero.

Therefore, the emf induced in the segment ab is $E_{ab}=-BV\left(bc\right)$.

(3) EMF induced on the loop

The emf force in the loop is,

$$\begin{gathered} E=E_{ab}+E_{bc}+E_{ca} \\ orE=-BV\left(bc\right)+BV\left(bc\right)+0 \\ orE=0 \end{gathered}$$

Therefore, the emf induced in the loop is zero.