Question

1.Find the force on a 115 m long wire at right angles to a 5.0 × 10–5 N/A ⋅ m magnetic field if the current through the wire is 400. A.

Answer 1

The formula for the Force on a current carrying wire is,

`\begin{gathered} F=IBlsin(\theta) \\ where, \\ F=Force \\ I=Current \\ B=Magnetic\text{ }Field \\ l=Length\text{ }Of\text{ }The\text{ }Wire \\ \theta=Angle\text{ }Between\text{ }Length\text{ }And\text{ }Magnetic\text{ }Field \end{gathered}`
`\begin{gathered} Here, \\ B=5*10^(-5)Tesla \\ l=115m \\ I=400A \\ \theta=90\mathring{} \\ So, \\ F=IBlsin(\theta)=400*5*10^(-5)*115* sin(90\mathring{\text{ }}) \\ So,\text{ }F=230000*10^(-5)=2.3N \end{gathered}`

So, Force on the wire is 2.3N