A conducting rectangular loop of area 5 cm\(^{2}\) and resistance 4 \(\Omega\) is removed from a region of uniform magnetic field, acting normal to the plane of the loop. The value of induced current I in the loop varies with time t, as shown in the figure.
Find:
(a) total charge that passed through the loop
(b) change in magnetic flux through the loop
(c) magnitude of magnetic field in the region
Show Hint
Charge is the area under the I-t graph. Then use Q = change in flux / R and set the change in flux equal to BA.
Step 1: Read the graph:
The graph shows a current that falls in a straight line from \(0.3\) A at \(t=0\) to zero at \(t=0.6\) s. The loop has area \(A=5\ \text{cm}^{2}=5\times10^{-4}\ \text{m}^{2}\) and resistance \(R=4\ \Omega\).
Step 2: Charge passed (part a):
Charge is the area under the \(I\)-\(t\) graph, \(Q=\int I\,dt\). This area is a triangle.
\[ Q=\frac{1}{2}\times\text{base}\times\text{height}=\frac{1}{2}\times0.6\times0.3=0.09\ \text{C} \]
Step 3: Change in flux (part b):
The induced emf is \(\varepsilon=-\dfrac{d\Phi}{dt}\), so the current is \(I=\dfrac{|\varepsilon|}{R}=\dfrac{1}{R}\left|\dfrac{d\Phi}{dt}\right|\).
Integrate over the whole time: \(\int I\,dt=\dfrac{\Delta\Phi}{R}\). So \(Q=\dfrac{|\Delta\Phi|}{R}\).
\[ |\Delta\Phi|=QR=0.09\times4=0.36\ \text{Wb} \]
Step 4: Magnetic field (part c):
The loop is taken completely out of the field. So the final flux is zero, and the change in flux equals the initial flux \(BA\) (the field is normal to the loop).
\[ B=\frac{|\Delta\Phi|}{A}=\frac{0.36}{5\times10^{-4}}=720\ \text{T} \]
Final Answer:
The charge is \(0.09\) C, the flux change is \(0.36\) Wb and the field is \(720\) T, using the values given in the question.
\[ \boxed{Q=0.09\ \text{C},\quad \Delta\Phi=0.36\ \text{Wb},\quad B=720\ \text{T}} \]
