Answer

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**Hint:**This question is similar to concept motional EMF in loop by generated we can solve this question by using formula $\left| e \right|=\dfrac{d\phi }{dt}$ and right hand thumb rule and for section (C) we will use the formula of the power $P={{I}^{2}}R$ to solve the question.

**Formula used:**

$\phi =B.A$

$\phi $ = flux

B =magnetic field

A = area

$\left| e \right|=\dfrac{d\phi }{dt}$

e = EMF

$\dfrac{d\phi }{dt}$ = change in the flux

$P={{I}^{2}}R$

P = power dissipation across resistance

I = current in the circuit

R = resistance

**Complete step by step solution:**

A).If the conductor is moving in positive x – direction with the velocity v what the change in the flux in one sec is.

$\to $ It is given that the time for displacement is

t = 1sec

We know that formula for the velocity

$v=\dfrac{displacement}{t}$

Hence

Displacement = $vt$

In time t distance travelled by the conductor is as shown in the figure

$vt......\left( 1 \right)$

Now change in the flux

$\left| e \right|=\dfrac{d\phi }{dt}$

We know that the flux

$\phi =B.A$

$\begin{align}

& \dfrac{d\phi }{dt}=\dfrac{d(BA)}{dt} \\

& =A\times \dfrac{dB}{dt}.....(1) \\

\end{align}$

$\to $ Now for the area we will use the length l and distance travelled vt as shown in the figure.

$\to $So area,

$A=lvt.....\left( 2 \right)$

$\to $ Now substitute value of the equation (2) in equation (1)

$\left| e \right|=lvt\times \dfrac{dB}{dt}$

$\to $ Here magnetic field is constant across circuit

$\begin{align}

& \left| e \right|=l\times v\times 1\times B \\

& \left| e \right|=Bvl.....(3) \\

\end{align}$

B) Magnitude and the direction of the current

$\to $ We know that,

$I=\dfrac{\left| e \right|}{R}.....\left( 4 \right)$

$\to $ Now substitute value of the equation (3) in equation (4)

$I=\dfrac{Bvl}{R}....(5)$

$\to $ Now to know that direction of the current we will use thumb rule as magnetic field vertically downwards current direction will be

C) Power dissipation in the resistor:-

$\to $Power dissipation in resistor is given by below formula

$P={{I}^{2}}R.....(6)$

Now substitute value of the equation (5) in equation (6)

$\begin{align}

& P={{\left( \dfrac{Bvl}{R} \right)}^{2}}R \\

& P=\dfrac{{{B}^{2}}{{v}^{2}}{{l}^{2}}}{R} \\

\end{align}$

**Note:**

When we find direction of the current using the thumb rule we have to be careful of the direction of the magnetic field because in some cases it is vertically upwards.

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