Question: In Figure, two straight conducting rails form a right angle. A conducting bar in contact with the rails starts at the vertex at time t = 0and moves with a constant velocity of 5.20m/salong them. A magnetic field with B = 0.350 Tis directed out of the page. (a) Calculate the flux through the triangle formed by the rails and bar atT = 3.00S. (b) Calculate the emf around the triangle at that time. (c) If the emf is$\epsilon =a{t}^n$, where a and n are constants, what is the value of n?

1. Velocity of conducting bar v = 5.20m/s
2. Magnetic field B = 0.350 T

Find the area of the triangle by referring to the figure 30.45. Use the area and evaluate the expression of the flux which involves the magnetic field, the velocity and the time. From Faraday’s law, the emf is induced in the loop when the number of the magnetic field lines that pass through the loop is changing.

Faraday's law of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Formulae are as follow:

$$\begin{gathered} \mathbf{\varphi }=\int \mathrm{B}.\mathrm{dA} \\ \mathrm{\epsilon }=\mathrm{N}\frac{\mathrm{d}\mathbf{\varphi }}{\mathrm{dt}} \end{gathered}$$

Where,$\mathrm{\varphi }$is magnetic flux, B is magnetic field, A is area, 𝜀 is emf, Nis number of turns.

To determine the flux,

$$\begin{gathered} \mathbf{\varphi }=\int \mathrm{B}.\mathrm{dA} \\ \mathbf{\varphi }=\int \mathrm{B}.\mathrm{dA}\mathrm{cos\theta } \\ \mathbf{\varphi }=\int \mathrm{B}.\mathrm{dA}\cos 0 \\ \mathbf{\varphi }=\mathrm{BA} \end{gathered}$$

By referring to the figure 30.45, the area of the triangle is changing due to the velocity of the conducting bar. The height of the triangle is changing with time and it is given as,

h = vt

By considering the geometry of the triangle, the area is given as,

$$\begin{gathered} A=\frac{1}{2}base\times height \\ A=\frac{1}{2}2h\times h \\ A=h^2 \\ A={\left(vt\right)}^2 \end{gathered}$$

Now, by substituting value to determine at ,

$$\begin{gathered} \mathbf{\varphi }=\int \mathrm{B}.\mathrm{dA} \\ \mathbf{\varphi }=\int v^2{t}^2 \\ \mathbf{\varphi }=\left(0.35T\right){\left(5.2m/s\right)}^2{\left(3.00s\right)}^2 \\ \mathbf{\varphi }=85.2Wb \end{gathered}$$

Hence, the flux through the triangle formed by the rails and bar at t = 3.00s is $\mathbf{\varphi }=85.2\mathrm{Wb}$.

The magnitude of emf induced around the triangle at.

$$\begin{gathered} \epsilon =-\frac{d\mathbf{\varphi }}{dt}and\mathbf{\varphi }=B{v}^2{t}^2 \\ \epsilon =-\frac{d}{dt}\left(B{v}^2{t}^2\right) \\ \epsilon =-2B{v}^{2}t \\ \epsilon =-2\left(0.35T\right)\left(5.2m/s\right){)}^2\left(3.0s\right) \\ \left|\epsilon \right|=56.784V\gg 56.8V \end{gathered}$$

Hence, the emf around the triangle at t = 3.00s is$\epsilon =56.8V$.

The value of n can be determined by comparing the above emf with $\epsilon =a{t}^n$.

It is observed that value of n = 1.

Hence, the value of n = 1.

Therefore, find the flux for the figure 30.45 by evaluating the area of the triangle formed by the conducting rail and the conducting bar. The flux is depending on the velocity of the conducting bar. The induced emf also can be determined from the flux.