Question

In a given circuit inductor of $ \mathrm{L} = 1\mathrm{mH}$ and resistance $ \mathrm{R} = 1\mathrm{\Omega }$ are connected in series to ends of two parallel conducting rods as shown. Now a rod of length $ 10 \mathrm{cm}$ is moved with constant velocity of 1 cm/s in magnetic field $ \mathrm{B} = 1\mathrm{T}.$ If rod starts moving at $ \mathrm{t} = 0$ then current in circuit after 1 millisecond is $ \mathrm{x} . {10}^{-3 }\mathrm{A}.$ Then the value of$ \mathrm{x} $is: (given $ {\mathrm{e}}^{-1} = 0.37$)

Answer 1

Step 1:Given data:

In a circuit, the inductance is given as $\mathrm{L}=1\mathrm{mH}$

The parallel-connected Resistance is given as $\mathrm{R}=1\mathrm{\Omega }$ , which is shown as:

The rod of length is given as= $10\mathrm{cm}$

Velocity is given as=$1\mathrm{cm}/\mathrm{s}$

Magnetic field $\mathrm{B}=1\mathrm{T}.$

The rod is given to start moving at $\mathrm{t}=0$ and the current in the circuit is measured after $=1\mathrm{millisecond}$

Current in the circuit $\left(\mathrm{i}\right)=?$

Step 2: Formula given for calculating the current in circuit :

The formula current in a given circuit can be given as:

$i=\frac{\overrightarrow{\epsilon }}{R}$

Where,

$\mathrm{R}=$Resistance of the circuit

And, $\overrightarrow{\epsilon }$ = emf (Electromotive force) of a circuit , which can be given as:

The formula for calculating the emf of a circuit can be given as:

$\overrightarrow{\epsilon }=(\mathrm{v}⃗\times \mathrm{B}⃗).\overrightarrow{\mathrm{dl}}$

$\mathrm{v}=$ the velocity of the circuit

$\mathrm{B}=$ Magnetic field

l= length of the rod

Step 2: Calculating the current in the circuit :

The emf is calculated as:

$$\begin{gathered} \overrightarrow{\epsilon }=(\mathrm{v}⃗\times \mathrm{B}⃗).\overrightarrow{\mathrm{dl}} \\ ={10}^{-2}\times 1\times {10}^{-3} \\ ={10}^{-3} \end{gathered}$$

Hence the current is given as:

$$\begin{gathered} i=\frac{{10}^{-3}(1-{\mathrm{e}}^{-1})}{1} \\ ={10}^{-3}(1-0.37) \\ =0.63\mathrm{mA} \end{gathered}$$

Hence, the value of$\mathrm{x}$is:$0.63.$