91影视

1. Length of rod L = 10 cm
2. eed of rod v = 5 m/s
3. Resistance of rod $R=0.400惟$
4. Current through long wire i = 100 A

e. Distance of the wire a = 10 mm

Find the expression of flux through the loop. By substituting this expression of flux in Faraday鈥檚 law, find the induced emf of the rod. Using Lenz鈥檚 law, find the direction of this induced emf. By substituting the value of emf in Ohm鈥檚 law, get the value of induced current of the loop. Using this current in the formula for rate of thermal energy generated, get the value of thermal energy generated in the loop. Using the magnetic force formula, find the force on the rod. Using the equation,find the rate of work due to the external agent.

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

Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant.

Lenz's law states that the current induced in a circuit due to a change in a magnetic field is directed to oppose the change in flux and to exert a mechanical force that opposes the motion

Formulae are as follow:

$$\begin{gathered} \left|蔚\right|=\frac{d{蠒}_B}{dt} \\ 蠒=BA \\ B=\frac{{渭}_{0}i}{2蟿蟿r} \\ P=i^{2}R \\ i=\frac{蔚}{R} \\ {F}_B=iLB \\ P=Fv \end{gathered}$$

Where,$桅$is magnetic flux, B is magnetic field, A is area, i is current, R is resistance, 饾渶 is emf, Nis number of turns, 饾渿₀is permeability, is the distance, L is length of rod, F is force, v is velocity, P is power.

Emf induced in the loop:

First, find the magnetic flux through the area enclosed by rod and rails.

Let x be the distance from the right end of the rails to the rod.

The magnetic field due the long straight current carrying wire at distance r is,

$B=\frac{{渭}_{0}i}{2蟿蟿r}$

Where,is the distance from the current carrying wire

Let鈥檚 consider the infinitesimal strip of length x and width dr, parallel to the current carrying wire and its distance is r from the wire.

So the area of the strip is dA = xdr

And the magnetic flux through this small area dA is,

$$\begin{gathered} d{蠒}_B=BdA=\frac{{渭}_{0}i}{2蟿蟿r}dA \\ d{蠒}_B=\frac{{渭}_{0}i}{2蟿蟿r}脳dr \end{gathered}$$

Integrating this equation, get the total flux through the area enclosed by rod and rails.

${蠒}_B=鈭�\frac{{渭}_{0}i}{2蟿蟿r}脳dr$

Here, the distance r varies from a + L

$$\begin{gathered} {蠒}_B=\frac{{渭}_{0}ix}{2蟿蟿}{鈭�}_a^{a+L}\frac{1}{r}dr \\ {蠒}_B=\frac{{渭}_{0}ix}{2蟿蟿}{\left[\ln r\right]}_a^{a+L} \\ {蠒}_B=\frac{{渭}_{0}ix}{2蟿蟿}\ln \left(\frac{a+L}{a}\right) \end{gathered}$$

Using Faraday鈥檚 law, the magnitude of emf induced in the loop is,

$$\begin{gathered} 蔚=\frac{d{蠒}_B}{dt} \\ 蔚=\frac{d}{dt}\left[\frac{{渭}_{0}ix}{2蟿蟿}\ln \left(\frac{a+L}{a}\right)\right] \end{gathered}$$

Taking constant terms out of the derivative,

$蔚=\frac{{渭}_{0}ix}{2蟿蟿}\ln \left(\frac{a+L}{a}\right)\frac{dx}{dt}$

Where,$\frac{dx}{dt}=v$ is the velocity of rod

$$\begin{gathered} 蔚=\frac{{渭}_{0}i}{2蟿蟿}\ln \left(\frac{a+L}{a}\right)v \\ 蔚=\frac{{渭}_{0}i}{2蟿蟿}\ln \left(\frac{a+L}{a}\right) \end{gathered}$$

$v=5m/s,L=10cm,a=10mm=1cm$

$$\begin{gathered} 蔚=\frac{\left(4蟿蟿脳{10}^{-7}{\displaystyle \frac{T.m}{A}}\right)\left(100A\right)\left(5m/s\right)}{2蟿蟿}\ln \left(\frac{1cm+10cm}{1cm}\right) \\ 蔚=2.40脳{10}^{-4}V \end{gathered}$$

Hence, emf induced in the loop is $2.40脳{10}^{-4}V$.

Current induced in the loop :

By using Ohm鈥檚 law, the current induced in the loop is,

$$\begin{gathered} i_l=\frac{蔚}{R} \\ {i}_l=\frac{2.40脳{10}^{-4}V}{0.400惟}=6脳{10}^{-4}A \end{gathered}$$

Since, the flux is increasing, the magnetic field produced by the induced current must be into the page in the region enclosed by the rod and rails. This means the current isClockwise.

Hence, current induced in the loop is$6脳{10}^{-4}A$.

Rate of thermal energy generated in the loop :

The rate of thermal energy generated is,

$$\begin{gathered} P=i_1^{2}R={\left(6脳{10}^{-4}A\right)}^2\left(0.400惟\right) \\ P=1.44脳{10}^{-7}W \end{gathered}$$

Hence, rate of thermal energy generated in the loop is$1.44脳{10}^{-7}W$

Magnitude of force applied to the rod to make it move at constant speed :

Since, the rod is moving with constant velocity, the net force on the rod is zero. To move the rod with constant velocity, the magnetic force and force due to the external agent which is moving the rod to the left must have the same magnitude, but opposite direction.

Let鈥檚 consider the infinitesimal segment of the rod with lengthand distance of this length segment from the current carrying wire is r.

Then the magnetic force on this small length element due to the current carrying rod is,

$d{F}_8=i_{l}dlB=i_{l}drB$

The magnetic field of current carrying wire is,

$$\begin{gathered} B=\frac{{渭}_{0}i}{2蟿蟿r} \\ d{F}_B=\left(\frac{{渭}_{0}i{i}_l}{2蟿蟿r}\right)dr \end{gathered}$$

Integrating this, the net magnetic force on the rod is,

$$\begin{gathered} F_B=\frac{{渭}_{0}i{i}_l}{2蟿蟿}{鈭�}_a^{a+L}\frac{1}{r}dr \\ {F}_B=\frac{{渭}_{0}i{i}_l}{2蟿蟿}\ln \left(\frac{a+L}{a}\right) \end{gathered}$$

$$\begin{gathered} F_B=\frac{\left(4蟿蟿脳{10}^{-7}{\displaystyle \frac{T.m}{A}}\right)\left(100A\right)\left(6脳{10}^{-4}A\right)}{2蟿蟿}\ln \left(\frac{1cm+10cm}{1cm}\right) \\ {F}_B=2.87脳{10}^{-8}N \end{gathered}$$

This force is towards the right and the external agent also applying the same magnitude of force but it is opposite and towards the left.

Hence, magnitude of force that must be applied to the rod to make it move at a constant speed is$2.87脳{10}^{-8}N$

Rate of work due to external agent :

Using the equations 7 - 48, the external agent does work at the rate,

$$\begin{gathered} P=Fv \\ P=\left(2.87脳{10}^{-8}N\right)脳\left(5m/s\right) \\ P=1.44脳{10}^{-7}W \end{gathered}$$

This is same as the thermal energy generated in the rod. Hence, all the energy supplied by the external agent to the rod is converted to thermal energy.

Hence,rate of work due to the external agent is$1.44脳{10}^{-7}W$

Therefore, using Faraday鈥檚 law, find the emf induced in the loop. Using Ohm鈥檚 law, find the current in the loop. Using the formula of thermal energy generated, find the thermal energy generated in the loop. Using the formula for magnetic force, find the force on the rod. Using the formula for power, find the rate of work.