A rectangular wire loop of sides \[8\,cm\] and \[2\,cm\] with a small cut is moving out of a region of uniform magnetic field of magnitude \[0.3\,T\] directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is \[1\,cm{s^{ - 1}}\] in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

Hint:We start by noting down the given information in the question. Then we move onto finding the area of the rectangle mentioned. We then find the emf of the given loop. This is found out along the breadth and along the length.

Formulas used:
Area of a rectangle with given length and breadth is given by the formula,
\[A = l \times b\]
Where \[l\] is the length of the rectangle and \[b\] is the breadth of the rectangle.
Emf of the loop is given by the formula,
\[e = Blv\]
Where \[B\] is the magnetic field associated with the loop and \[v\] is the velocity of the loop.
The time taken for a given value of velocity and distance is,
\[t = \dfrac{S}{v}\]
Where \[S\] is the distance travelled in the given time and \[v\] is the velocity.

Complete step by step answer:
Let us start by writing down the given values one by one,
Length of the given rectangular loop is, \[l = 8\,cm = 0.08\,m\].
Breadth of the given rectangular loop is, \[b = 2\,cm = 0.02\,m\].
The value of the magnetic field associated with the given rectangular loop, \[B = 0.3T\].
The velocity at which the rectangular loop moves is,
\[v = 1\,cm{s^{ - 1}} = 0.01\,m{s^{ - 1}}\]

(a) Now we move onto finding the area of the rectangular loop using the formula,
\[A = l \times b\]
We get,
\[A = l \times b \\
\Rightarrow A = 0.02 \times 0.08 \\
\Rightarrow A = 0.0016 \\
\Rightarrow A = 16 \times {10^4}{m^2}\]

Now we move onto finding the emf of the loop along the breadth, which is given by the formula, \[e = Blv\]. We substitute the value and get,
\[e = Blv \\
\Rightarrow e = 0.3 \times 0.08 \times 0.01 \\
\Rightarrow e = 2.4 \times {10^{ - 4}}V\]
The time taken for this to travel along the width or breadth is given by dividing the distance travelled and the velocity of the rectangular loop. That is,
\[t = \dfrac{S}{v} \\
\Rightarrow t = \dfrac{{0.02}}{{0.01}} \\
\therefore t = 2\,s\]

Hence the induced voltage of \[2.4 \times {10^{ - 4}}V\] lasts upto a full \[2s\].

(b) Now we do the exact same thing along the length of the rectangular loop,
The emf of the loop along the breadth, which is given by the formula, \[e = Blv\]
\[e = Blv \\
\Rightarrow e = 0.3 \times 0.02 \times 0.01 \\
\Rightarrow e = 6 \times {10^{ - 5}}\]
We move onto finding the time taken for this, we do the same procedure we did for finding the time taken and arrive at,
\[t = \dfrac{S}{v} \\
\Rightarrow t = \dfrac{{0.08}}{{0.01}} \\
\therefore t = 8\,s\]

Hence the induced voltage of \[6 \times {10^{ - 5}}V\] lasts upto a time of \[8s\].

Note: Electromotive force or emf is defined as the electric potential produced by an electrochemical cell or by changing the magnetic field. Induced emf is produced whenever there is a change in the magnetic field around a conductor. The induced emf is the negative of the rate of the change of magnetic flux times the number of coils.