Question

The square loop shown in$FIGUREP30.53$ moves into a $0.80\mathrm{T}\mathrm{ALC}$magnetic field at a constant speed of$10m/s$.The loop has a resistance of $0.10\mathrm{\Omega }$, and it enters the field at $t=0\mathrm{s}.$

a. Find the induced current in the loop as a function of time. Give your answer as a graph of ii versus $t$from $t=0\mathrm{s}$to $t=0.020\mathrm{s}.$

b. What is the maximum current? What is the position of the loop when the current is maximum?

Short answer

Part $\left(a\right)$

$\left(a\right)$The loop's induced current as a respect to time is$I_{\text{induced}}=\left(1.6\times {10}^3\right)t.$

Part $\left(b\right)$

$\left(b\right)$When the loop is half-way through the magnetic field, the maximum current $11A$is generated.

Step-by-step solution

Step: 1  Find induced current (part a) 

The magnetic flux $\varphi$is the amount of magnetic field that passes through a loop of area $A$. The magnetic flux is provided by when the magnetic field creates an angle with the plane.

${\mathrm{\Phi }}_{\mathrm{m}}=BA$

In time $t$, the loop travels at speed $v$for a distance $x$. As a result, the distance is $x=vt$. When the entire loop is put in the magnetic field, this distance indicates the loop's side. As a result, the loop's area will be

$A=x^2=v^2{t}^2$

The induced emf, as defined by Faraday's law, is the change in magnetic flux inside the loop, and it is provided by equation

$\epsilon =\frac{d{\mathrm{\Phi }}_{\mathrm{m}}}{dt}=B\frac{dA}{dt}=B\frac{d}{dt}\left(v^2{t}^2\right)=2v^{2}Vt.$

We utilise Ohm's law to determine the induced current $I$via the inner loop, as indicated in the next equation.

$$\begin{gathered} I_{\text{induced}}=\frac{\epsilon }{R}=\frac{2v^{2}Bt}{R} \\ \begin{array}{l}{I}_{\text{induced}}=\frac{2v^{2}Bt}{R}\\ I_{\text{induced}}=\frac{2(10\mathrm{m}/\mathrm{s}{)}^2(0.80\mathrm{T})}{\left(0.10\mathrm{\Omega }\right)}t\\ I_{\text{induced}}=\left(1.6\times {10}^3\right)t.\end{array} \end{gathered}$$

Step: 2 Finding value of t: (part a)

In the range of $0s$to $0.020s$, we wish to draw the induced current versus time $t$distance$vt$. As the loop moves in the magnetic field, the current increases with time, as indicated by our result. When the loop contains a magnetic field, the magnetic field inside the loop remains constant, and the current becomes zero. As a result, when the magnetic field covers half of the loop, the current achieves its maximum amount. To fill half of the loop, the loop moves a distance , which is computed using the Pythagorean theorem.

$\begin{array}{l}(vt{)}^2+(vt{)}^2=(10\mathrm{cm}{)}^2\\ vt=\sqrt{\frac{100{\mathrm{cm}}^2}{2}}\\ t=\sqrt{\frac{100{\mathrm{cm}}^2}{2}\frac{1}{v}}\end{array}$

We use $v=10m/s$to retrieve $t$,

$$\begin{gathered} t=\sqrt{\frac{100{\mathrm{cm}}^2}{2}}\frac{1}{v} \\ t=\sqrt{\frac{100{\mathrm{cm}}^2}{2}}\left(\frac{1}{10\times {10}^2\mathrm{cm}/\mathrm{s}}\right) \\ t=7\times {10}^{-3}\mathrm{s}=7\mathrm{ms}. \end{gathered}$$

Step: 3 Find the maximum induced current: (part a)

Now, multiply the value of $t$by the maximum induced current.

$$\begin{gathered} I_{\text{induced}}=\left(1.6\times {10}^3\right)\left(7\times {10}^{-3}\mathrm{s}\right) \\ {I}_{\text{induced}}=11\mathrm{A}. \end{gathered}$$

To make the graph, we draw a straight line from $t=7$to $11A$, then drop to zero after $7\mathrm{ms}$at $\left(14\mathrm{ms}\right)$and keep it zero until $t=20\mathrm{ms}=0.020\mathrm{s}$, as indicated in the diagram.

Step: 4 Finding maximum current (part b) 

The magnetic field's intensity is highest at the loop's centre and drops as you go away from it. As the radius of the circular loop is increased, the magnetic field intensity diminishes. With an increase in electric current passing through the circular wire loop, the magnetic field intensity increases.The highest current $11A$is created when the loop is half-way through the magnetic field.