# Doubt from Electromagnetic Induction

**Achyut_2021**#1

In case of a simple harmonic oscillator, the speed of the moving mass is given by : v(t)=Acos(\omega_0t) where \omega_0 is the natural frequency of oscillation.

We know that : V_m=[\vec{B}\space\vec{l}\space\vec{v}], where [.] represents the box or scalar triple product. Expanding the above product, we get : V_m=\vec{B}.(\vec{l}\space X\space\vec{v}).

Taking the direction of \vec{v} along the line segment PQ, we can break the whole motion into four parts :

\to *Part 1*: Rod going from mean to extreme position towards right.

In this case we know that v is maximum at the starting point and varies as cosine, so the graph will be like a cosine only. We see that \vec{v}\space X\space\vec{l} points outwards here and hence its dot product with the magnetic field will be positive only and hence the bar will trace the quarter of a cosine wave till it reaches the end point where the velocity vanishes.

\to *Part 2*: Rod going from right extreme to mean position towards left.

In this case, initially the motional emf is zero as the speed is zero. The direction of \vec{v}\space X\space\vec{l} will be inwards here and hence its dot product with the magnetic field will be negative. Here, as the rod reaches the initial position the motional emf will reach maxima but with opposite sign and hence the remaining part of the cosine wave will be reached.

\to *Part 3*: Rod going from mean to extreme position towards left.

We can clearly see that the scalar triple product is positive and hence there will be an abrupt jump in the motional emf to its maximum value. And hence, same curve as part 1 will be traced.

\to *Part 2*: Rod going from left extreme to mean position towards right.

After the above discussion, its quite clear now that the remaining portion of the graph will be as in part 2.

Thus B is the correct answer.

Do let me know if you have any doubt.