91Ó°ÊÓ

We have a conducting rod of length L = 30.0 cm which moves in a magnetic field with a magnitude of B = 0.450 T, and points into the page as shown in the following figure.

a) We know that a conductor moving in a magnetic field may have a potential difference induced across it, depending on how it is moving, the magnitude of the induced emf is given by,

$\mathrm{Î\mu }=vBL\sin (\mathrm{Ï•})$

Where$\mathrm{ϕ}$is the angle between the velocity and the magnetic field. In our case, this angle is $\mathrm{ϕ}$= 90.0°, substitute with the givens we get,

$$\begin{gathered} \mathrm{Î\mu }=(5.00m/s)(0.450T)(0.300m)\left(sin\right({90}^{\mathrm{ο}}\left)\right)=0.675V \\ \mathrm{Î\mu }=0.675V \end{gathered}$$

(b) According to the right-hand rule, the positive charges are moved to end b, so b is at the higher potential.

(c) The magnitude of the electric field is,

$$\begin{gathered} E=\frac{V}{L}=\frac{0.675V}{0.300m}=2.25V/m \\ E=2.25V/m \end{gathered}$$

we know that the electric field is radially outward from a positive charge and radially in toward a negative point charge, therefore the direction of the electric field is from b to a.

(d) As shown in part (b), side b has an excess of a positive charge.

(i) If the rod moves parallel to ab and it has no appreciable thickness L =0, so the emf is zero, and

(ii) if the rod is directly out of the page, the emf is zero, since it is parallel to the magnetic field, hence no magnetic force acts on the charges.