Question

A wire carries a current of 6.52 A in a direction that makes an angle of 44.8 degrees with the direction of a magnetic field of strength 0.258 T. What is the magnetic force on a 3.61 m length of wire

Answer 1

`F=4.27N`

Step-by-step explanation:

From the question we are told that:

Current
`I = 6.52 A`

Angle
`\theta= 44.8 \textdegree`

Magnetic field
`B= 0.258 T.`

Length
`l=3.61m`

Generally the equation for Magnetic force F is mathematically given by

`F=BIlsin\theta`

Therefore

`F=0.258*6.52*3.61*sin 44.8`

`F=4.27N`