The force on a wire carrying 8.75 A is a maximum of 1.28 N when placed between the pole faces of a magnet

If the pole faces are 55.5 cm in diameter, what is the approximate strength of the magnetic field?

Respuesta :

Answer:

Explanation:

Given

Current in the conductor [tex]i=8.75 A[/tex]

maximum Force [tex]F=1.28 N[/tex]

If Poles Faces are 55.5 cm in diameter i.e. Length [tex]L=55.5 cm[/tex]

Force experience by a current carrying conductor is given by

[tex]F=Bil[/tex]

where [tex]B=magnetic\ Field[/tex]

[tex]i=current [/tex]

[tex]F=force[/tex]

[tex]1.28=B\times 8.75\times 0.555[/tex]

[tex]B=0.263 T[/tex]

thus Strength of Magnetic Field is 0.263 T