Question

A long straight copper wire carries current in an upward direction. If a charge of 25 kC passes at a point within 0.5 s, determine the magnitude of the current induced at the point and the magnitude and direction of the magnetic field produced at point P where point P is 275 cm to the right of the wire.

Answer 1

The current induce at the point can be given as,

`I=(q)/(t)`

Plug in the known values,

`\begin{gathered} I=\frac{(25\text{ kC)(}\frac{1000\text{ C}}{1\text{ kC}})}{0.5\text{ s}}(\frac{1\text{ A}}{1\text{ C/s}}) \\ =50000\text{ A} \end{gathered}`

The magnetic field produced at the point P is given as,

`B=(\mu_0I)/(2\pi r)`

Substituting values,

`\begin{gathered} B=\frac{(4\pi*10^(-7)\text{ Tm/A)(50000 A)}}{2\pi(275\text{ cm)(}\frac{1\text{ m}}{100\text{ cm}})} \\ \approx3.63*10^(-3)\text{ T} \end{gathered}`

Therefore, the current induced in the wire is 50000 A, the magnetic field acting on the point is

`3.63*10^(-3)\text{ T}`

which acts in the anticlockwise direction as the current moves in the upward direction.