An electromagnet produces a magnetic field of 0.550 T in a cylindrical region of radius 2.50 cm between its poles. A straight wire carrying a current of 10.8 A passes through the centre of this region and is perpendicular to both the axis of the cylindrical region and the magnetic field. What magnitude of force does this field exert on the wire?

It is given and a straight wire carrying a current of l = 10.8 A . That passes through the centre of the region and is perpendicular to both the axis of the cylindrical magnetic field.

The magnetic influence on moving electric charges, electric currents, and magnetic materials is described by a magnetic field, which is a vector field. In a magnetic field, a moving charge experiences a force that is perpendicular to both its own velocity and the magnetic field.

As the angle between the length l and the magnetic field is $\mathrm{\Phi }=90°$, and the length of the wire in magnetic field is l = 2r = 0.050 m .

Hence the magnitude of the force is therefore:

$F=llB\sin \varphi$

Here I is the current, l is the wire length and B is the magnetic field.

Substitute all the value in the above equation.

$\begin{array}{r}\mathrm{F}=\mathrm{IIBsin}\mathrm{\varphi }\\ =\mathrm{IIBsin}{90}^{\circ }\\ =10.8\mathrm{A}\times 0.05\mathrm{m}\times 0.55\mathrm{T}\\ =0.297\mathrm{N}\end{array}$

Therefore, the force is 0.297 N