Question

A wire loop is placed in a region of time varying magnetic field which is oriented orthogonally to the plane of the loop as shown in the figure. The graph shows the magnetic field variation as the function of time. Assume the positive emf is the one which drives a current in the clockwise direction and seen by the observer in the direction of B shown. Which of the following graphs best represents the induced emf as a function of time.​

Answer 1

Answer:

The correct graph is Option - (3)

Explanation:

We are given, that a wire loop is placed in a region of magnetic field which is varying w.r.t time. And, we need to plot a graph which represents induced emf as function of time.

From the formula we know that,

$\longrightarrow\sf \phi = B.A$

( Dividing it with time and limiting it to zero )

$\longrightarrow\sf \dfrac{d\phi}{dt}= A\;.\;\dfrac{dB}{dt}$

[ Multiplying with negative sign on both sides ]

$\longrightarrow\sf - \dfrac{d\phi}{dt}= - \;A\;.\;\dfrac{dB}{dt}\\\\\\\\\longrightarrow\sf \varepsilon = - \;A\;.\;\dfrac{dB}{dt} \ \ \ \ \ \ \ \ \ \ \because \Bigg[\varepsilon= - \dfrac{d\phi}{dt}\Bigg]$

From the graph (In the attachment) we can make 3 cases.

Case-1

For region OP ie ( o ≤ t ≤ t₁ ) dB/dt is positive, thus the emf will be negative. As ε = -A . dB/dt.

Case-2

For region PQ ie ( t₁ ≤ t ≤ t₂ ) dB/dt is zero, thus the emf will be zero. As ε = -A . dB/dt.

Case-3

For region QRS ie ( t₂ ≤ t ≤ t₃ ) dB/dt is negative, thus the emf will be Positive. As ε = -A . dB/dt.

So, after plotting it, we get the option - 3 is correct. (Refer the attachment).