Question

A wire carries a steady current of 2.80 A. A straight section of the wire is 0.750 m long and lies along the x axis within a uniform magnetic field, = 1.50 T. If the current is in the positive x direction, what is the magnetic force on the section of wire?

Answer 1

The magnetic force in the wire is 3.15N

Step-by-step explanation:

Given

current I= 2.80 A.

length of conductor L= 0.75 m

Magnetic field, B = 1.50 T

∅=90

according to Fleming's left hand rule the conductor will observe a force perpendicular to it

Applying the formula

`F= BIL* sin(90)`

`F=1.50* 2.80* 0.750* sin(90)\\\F= 3.15N`

Note: sine(90)= 1