A wire 0.80m long carrying a current of 3.2A is perpendicular to a 0.15T magnetic field (B field). What

is the force on the wire? (Use F=BIL to solve!)

Input answer with 3 decimal places (for example: 555) and do NOT include the units!

E

1.344 N

Explanation:

The expression for the for on a wire carrying current in a magnetic field is given as,

F = BILsinФ Equation 1

Where F = Force on the wire, B = magnetic field, I = current, L = Length of the wire, Ф = angle between the conductor and the magnetic field.

Given: B = 0.15 T, I = 3.2 A, L = 0.8 m, Ф = 90° (perpendicular)

Substitute into equation 1

F = 0.15(3.2)(0.8)(sin90°)

F = 1.344 N

Hence the Force on the wire = 1.344 N

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm