Question

The conducting rod abshown in Fig. makes contact with metal rails caand db. The apparatus is in a uniform magnetic field of 0.800 T, perpendicular to the plane of the figure.

(a) Find the magnitude of the emf induced in the rod when it is moving toward the right with a speed of 7.50 m/s.

(b) In what direction does the current flow in the rod?

(c) If the resistance of the circuit abdcis 1.50 $\mathbf{\Omega }$(assumed to be constant), find the force (magnitude and direction) required to keep the rod moving to the right with a constant speed of 7.50 m/s. You can ignore friction.

(d) Compare the rate at which mechanical work is done by the force (Fv) with the rate at which thermal energy is developed in the circuit ( ${\mathit{I}}^{\mathit{2}}\mathit{R}$).

Short answer

1. The magnitude of the emf induced in the rod when it is moving toward the right with a speed of 7.50 m/s is 3.00V.
2. The current flow in the rod in the b to a direction.
3. The force required to keep the rod moving to the right with a constant speed of 7.50 m/s is equal to 0.800N in the right direction.
4. The rate at which mechanical work is done by the force (Fv) and the rate at which thermal energy is developed in the circuit ($I^{2}R$) are equal

Step-by-step solution

Calculate the magnitude of the emf and direction of the current.

We have a conducting rod ab, which makes contact with metal rails ca and db, the whole device is placed perpendicularly in a magnetic field of B = 0.800 T, as shown in the following figure.

(a) We need to find the magnitude of the emf induced in the rod when it is moving toward the right with a speed of v = 7.50 m/s as shown in the figure. Let be the length of the expanding side db, and L is the length of the constant length side ab. Then, the magnetic flux through the loop abcd is,

${\varphi }_B=BA=BLx$

the magnitude of the induced emf is, therefore,

substitute with the givens we get,

$\begin{array}{l}\mathrm{\epsilon }=(7.50\mathrm{m}/\mathrm{s})\left(0.800\mathrm{T}\right)\left(0.500\mathrm{m}\right)=3.00\mathrm{V}\\ \mathrm{\epsilon }=3.00\mathrm{V}\end{array}$

(b) The area of the loop abcd increases as the bar moves to the right, hence the magnetic flux, and the external magnetic field points into the page, so the induced magnetic field must point out of the page (according to the Lenz's law), so the induced current must circulate counterclockwise, which means from b to a in the rod.

Calculate force, mechanical work, and thermal energy.

(c) if the resistance of the circuit abdc is R =1.50 , then the induced current is,

$$\begin{gathered} \mathrm{I}=\frac{\mathrm{\epsilon }}{\mathrm{R}}=\frac{3.00\mathrm{V}}{1.50\mathrm{\Omega }}=2.00\mathrm{A} \\ \mathrm{I}=2.00\mathrm{A} \end{gathered}$$

the force acting on the bar is, therefore,

$$\begin{gathered} {\mathrm{F}}_1=\mathrm{IBLsin}\left(\mathrm{\varphi }\right) \\ =\left(2.00\mathrm{A}\right)\left(0.500\mathrm{m}\right)\left(0.800\mathrm{T}\right)\sin \left(90°\right) \end{gathered}$$

according to the right-hand rule this force point to the left. To keep the bar moving to the right at a constant speed, the net force acting on the rod must be zero, therefore, an external force of ${\mathrm{F}}_{\mathrm{ext}}=0.800\mathrm{N}$must be applied to the bar to the right

(d) The rate at which work is done by the external force is,

$P_{mech}=F_{ext}v=\left(0.800N\right)\left(7.50m/s\right)=6.00W$

and the rate at which thermal energy is developed in the circuit is,

${\mathrm{P}}_{\mathrm{therm}}={\mathrm{I}}^2\mathrm{R}={\left(2.00\mathrm{A}\right)}^2\left(1.50\mathrm{\Omega }\right)=6.00\mathrm{W}$

the two rates are equal.