The magnetic force on a current-carrying wire is given by:
F = iL×B
F = magnetic force, i = current, L = wire length vector, B = magnetic field
Note we are taking the cross product of the iL and B vectors, not the product of two scalars.
A) Given values:
i = 12A
L = AB = (2, 1, 1)m - (1, 1, 1)m = <1, 0, 0>
B = <-2, 3, 4>10⁻³T
Plug in and solve for F:
F = 12<1, 0, 0>×10⁻³<-2, 3, 4> = 10⁻³(<12, 0, 0>×<-2, 3, 4>)
Beware, we'll use array notation to show the cross product calculation:
F = 10⁻³[tex]\left[\begin{array}{ccc}i&j&k\\12&0&0\\-2&3&4\end{array}\right][/tex]
F = 10⁻³<(0)(4)-(0)(3), (0)(-2)-(12)(4), (12)(3)-(0)(-2)>
F = 10⁻³<0, -48, 36>N
B) Given values:
i = 12A
L = AB = (3, 5, 6)m - (1, 1, 1)m = <2, 4, 5>
B = <-2, 3, 4>10⁻³T
Plug in and solve for F:
F = 12<2, 4, 5>×10⁻³<-2, 3, 4> = 10⁻³(<24, 48, 60>×<-2, 3, 4>)
F = 10⁻³[tex]\left[\begin{array}{ccc}i&j&k\\24&48&60\\-2&3&4\end{array}\right][/tex]
F = 10⁻³<(48)(4)-(60)(3), (60)(-2)-(24)(4), (24)(3)-(48)(-2)>
F = 10⁻³<12, -216, 168>N
