Question

A 3.0 cm wire carrying a current of 10A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27T . what is the magnetic force on the wire ?

Answer 1

Given:

• Length of wire = 3.0 cm

,

• Current = 10 A

,

• Magnetic field = 0.27 T

Let's find the magnetic force on the wire.

To find the magnetic force on the wire, apply the formula:

`F=BILsin\theta`

Where:

B is the magnetic field = 0.27 T

I is the current = 10 A

L is the length of wire in meters = 3.0 cm = 0.03 m

Since it is perpendicular, θ = 90 degrees.

F is the magnetic force.

Substitute the values into the formula and solve for F.

We have:

`\begin{gathered} F=0.27*10*0.03sin90 \\ \\ F=0.27*10*0.03(1) \\ \\ F=0.081\text{ N} \\ \\ F=8.1*10^(-2)N \end{gathered}`

Therefore, the magnetic force on the wire is 8.1 x 10⁻² N.

ANSWER:

8.1 x 10⁻² N