Question

in the figure, a conducting rod of length ( l = 29.0 mathrm{~cm} ) moves in a magnetic field ( vec{b} ) of magnitude ( 0.440 mathrm{~t} ) directed into the plane of the figure. the rod moves with speed ( v = 4.80 mathrm{~km} / mathrm{s} ) in the direction shown. (figure 1 of 1) part c when the charges in the rod are in equilibrium, what is the magnitude ( e ) of the electric field within the rod? express your answer in volts per meter to at least three significant figures. view available hint(s) ( e= ) ( mathrm{v} / mathrm{m} ) part d which point, a or b, has a higher potential? view available hint(s)

Explanation:

Step1: Use the formula for equilibrium of forces

When charges are in equilibrium, \( qE = qvB \). So, \( E = vB \).

Step2: Convert the velocity unit

Given \( v = 4.85\ \text{km/s}=4850\ \text{m/s}\), \( B = 0.446\ \text{T}\).

Step3: Calculate the electric field

Substitute \( v = 4850\ \text{m/s}\) and \( B = 0.446\ \text{T}\) into \( E = vB \).
\( E=(4850\ \text{m/s})\times(0.446\ \text{T}) = 2163.1\ \text{V/m}\approx2160\ \text{V/m}\)

Using the right - hand rule for the magnetic force on positive charges. Positive charges experience a force. The direction of the magnetic force on positive charges in the rod: \(F = q\vec{v}\times\vec{B}\). With \(v\) (direction of motion of the rod) and \(B\) (into the page), positive charges move towards end \(a\). Electric field is from \(a\) to \(b\). Since electric potential increases in the direction opposite to the electric field (for positive charges), point \(a\) has a higher potential.

Answer:

\(2160\ \text{V/m}\)

For Part D: