Question

An electric bus operates by drawing direct current from two parallel overhead cables, at a potential difference of 600 V, and spaced 55 cm apart. When the power input to the bus’s motor is at its maximum power of 65 hp, (a) what current does it draw and (b) what is the attractive force per unit length between the cables?

Short answer

a) The current drawn by the electric bus is 80.8 A .

b) The attractive force per unit length between the cables is $2.4\times {10}^{-3}$N/m.

Step-by-step solution

(a) Determination of the current drawn by the electric bus

The expression relating power, voltage and current is,

P = VI

The power output in this case is 65 hp which is equal to 48,490 W.

Rearrange and solve for current,

$\begin{array}{r}\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}\\ =\frac{48,490\mathrm{W}}{600\mathrm{V}}\\ =80.8\mathrm{A}\end{array}$

Thus, the current drawn by the electric bus is 80.8 A.

(b) Determination of the attractive force per unit length between the cables

The expression for attractive force between two current carrying wires is,

$F=\frac{{\mu }_{0}L{I}_1{I}_2}{2\pi r}$

The force per unit length is,

$\begin{array}{r}\frac{\mathrm{F}}{\mathrm{L}}=\frac{{\mathrm{\mu }}_0{\mathrm{I}}_1{\mathrm{I}}_2}{2\mathrm{\pi r}}\\ =\frac{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}\right)(80.8\mathrm{A}{)}^2}{2\mathrm{\pi }(0.55\mathrm{m})}\\ =2.4\times {10}^{-3}\mathrm{N}/\mathrm{m}\end{array}$

Thus, the force of attraction is $2.4\times {10}^{-3}\mathrm{N}/\mathrm{m}$.