In Figure, a metal rod is forced to move with constant velocity $\overrightarrow{\mathbf{v}}$along two parallel metal rails, connected with a strip of metal at one end. A magnetic field of magnitude B = 0.350 T points out of the page.(a) If the rails are separated by L=25.0 cmand the speed of the rod is 55.0 cm/s, what emf is generated? (b) If the rod has a resistance of 18.0$\mathit{\Omega }$and the rails and connector have negligible resistance, what is the current in the rod?

(c) At what rate is energy being transferred to thermal energy?

1. Magnetic field B = 0.350 T
2. Separation of rails L = 25 cm
3. Speed of rod v = 55 cm/s
4. Resistance of rod$R=18\Omega$

Use the concept of motional emf to find the emf induced in the rod. Using Ohm’s law, find the current in the rod. Substituting the value of current and resistance in the equation for power, find the rate of transfer of thermal energy of the rod.

Faraday's law 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.

Formulae are as follow:

$$\begin{gathered} \epsilon =BLv \\ i=\left|\frac{\epsilon }{R}\right| \\ P=i^{2}R \end{gathered}$$

Where, $\varphi$is magnetic flux, B is magnetic field, i is current, R is resistance, 𝜀 is emf, v is velocity, L is separation.

The emf generated:

When the rod is moving with velocity v in the uniform magnetic field, the emf is induced in the rod.

Using equation 30-8, ,

$\epsilon =BLv$

By substituting the given values,

$$\begin{gathered} \epsilon =\left(0.350T\right)\left(0.250m\right)\left(0.55m/s\right) \\ \epsilon =0.0481V \end{gathered}$$

Hence, the emf generated is 0.0481 V.

Current induced in rod :

Using Ohm’s law, the induced current in the rod is given by,

$$\begin{gathered} i_{rod}=\left|\frac{\epsilon }{R}\right| \\ {i}_{rod}=\frac{0.0481V}{18\Omega }=0.00267A \end{gathered}$$

Hence, the current induced in rod is 0.00267 A.

Rate of transfer of thermal energy :

The rate of transfer of thermal energy is given by,

$P=i^{2}R$

$P={\left(0.00267\right)}^2\left(18\Omega \right)=0.000129W$

Hence, therate of transfer of thermal energy is 0.000129 W.

Therefore, use the concept of motional emf to find the emf induced in the rod and Ohm’s law to the current in the rod and rate of transfer of energy.