Question

The field B = −2ax + 3ay + 4az mT is present in free space. Find the vector force exerted on a straight wire carrying 12 A in the aAB direction, given A(1, 1, 1) and:
(a) B(2, 1, 1);
(b) B(3, 5, 6).

Answer 1

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$\left[\begin{array}{ccc}i&j&k\\12&0&0\\-2&3&4\end{array}\right]$

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$\left[\begin{array}{ccc}i&j&k\\24&48&60\\-2&3&4\end{array}\right]$

F = 10⁻³<(48)(4)-(60)(3), (60)(-2)-(24)(4), (24)(3)-(48)(-2)>

F = 10⁻³<12, -216, 168>N