The L-shaped conductor in $FIGUREP30.54$ moves at $10m/s$across and touches a stationary $L$-shaped conductor in a $0.10T$magnetic field. The two vertices overlap, so that the enclosed area is zero, at $t=0s$. The conductor has a resistance of $0.010$ohms per meter.

a. What is the direction of the induced current?

b. Find expressions for the induced emf and the induced current as functions of time.

c. Evaluate$\in$and$l$at$t=0.10s$.

According to Lenz's law, the induced emf forces current through a wire loop in the reverse direction of the change in magnetic flux that creates the $emf$. To determine the direction of the induced current and the induced magnetic field, use the right-hand rule, where your enclosing four fingers represent the induced magnetic field and your thumb represent the current.

There is an induced magnetic field into the page because the applied magnetic field has to be in the page and its strength develops. This induced magnetic field causes current to flow in a clockwise direction.

The magnetic flux $\mathrm{\Phi }$is the amount of magnetic flux that flows through a loop of area $A$. The magnetic flux is provided by when the magnetization makes an angle with the plane.

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

The loop runs at localid="1648915127867" $v$per minute over a distance of localid="1648915131114" $x$in time localid="1648915134505" $t$in the direction of localid="1648915137366" $\theta ={45}^{\circ }$, localid="1648915140154" $x=(v\cos \theta )t$is the range. When the entire loop is inserted in the magnetic field, this distance represents the loop's side. As a result, the portal's area will be

$A=x^2=(v\cos \theta {)}^2{t}^2=v^2{t}^2\cos \theta$

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

localid="1648915144573" $\epsilon =\frac{d{\mathrm{\Phi }}_{\mathrm{m}}}{dt}=B\frac{dA}{dt}=B\frac{d}{dt}\left(v^2{t}^2{\cos }^2\theta \right)=2B\left(v^2{\cos }^2\theta \right)t.$

Put the values of $v,B$and $\theta$into equation by

$\begin{array}{l}\epsilon =2B\left(v^2{\cos }^2\theta \right)t\\ \epsilon =2\left(0.10\mathrm{T}\right)(10\mathrm{m}/\mathrm{s}{)}^2{\cos }^2{45}^{\circ }t\\ \epsilon =10t.\end{array}$

The loop resistivity $R$by

$\begin{array}{l}R=(0.010\mathrm{\Omega }/\mathrm{m})\left(4x\right)\\ R=(0.010\mathrm{\Omega }/\mathrm{m})(4v\cos \theta t)\\ R=(0.010\mathrm{\Omega }/\mathrm{m})4\left(10\mathrm{m}/\mathrm{scos}{45}^{\circ }\right)t\\ R=0.28t.\end{array}$

The induced current through inner loop by ohm's law is

$I_{\text{induced}}=\frac{\epsilon }{R}=\frac{10t}{R}=\frac{10t}{0.28t}=35\mathrm{A}.$

The induced $emf$at time $t=0.10s$and the current will be same.

$\begin{array}{l}\epsilon =10t=10\left(0.10\mathrm{s}\right)=1\mathrm{V}\\ I_{\text{induced}}=35\mathrm{A}.\end{array}$