# The current in a conductor is found to be moving from (-1, 2, -3) m to (-4, 5, -6) m. Given that the

The current in a conductor is found to be moving from (-1, 2, -3) m to (-4, 5, -6) m. Given that the force the conductor experiences is $\stackrel{\to }{F}=12\left({a}_{y}+{a}_{z}{\right)}^{N}$ and the magnetic field present in the region is $\stackrel{\to }{B}=2{a}_{x}T$, determine the current through the conductor. a.) -1 A;
b.) -2 A;
c.) This problem has no solution.
d.) 2 A;
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

partyjnopp9wa
Length of the conductor $L=\left(-4,-1\right){a}_{x}+\left(5,2\right){a}_{y}+\left(6,-3\right){a}_{z}=-3{a}_{x}+3{a}_{y}-3{a}_{z}$
We know that
magnetic force $F=I\left(l×B\right)$
$l×B=\left(-3{a}_{x}+3{a}_{y}-3{a}_{z}\right)×2{a}_{x}\phantom{\rule{0ex}{0ex}}=\left(-3{a}_{x}×2{a}_{x}\right)+\left(3{a}_{y}×2{a}_{x}\right)-\left(3{a}_{z}×2{a}_{x}\right)\phantom{\rule{0ex}{0ex}}=0+-6{a}_{z}-6{a}_{y}\phantom{\rule{0ex}{0ex}}=-6\left({a}_{y}+{a}_{z}\right)$
Substituting this in equ
we get
$12\left({a}_{y}+{a}_{z}\right)=I×-6\left({a}_{y}+{a}_{z}\right)\phantom{\rule{0ex}{0ex}}I=\frac{12\left({a}_{y}+{a}_{z}\right)}{-6\left({a}_{y}+{a}_{z}\right)}=-2A\phantom{\rule{0ex}{0ex}}I=-2A$
Step 2
Current I=-2 A
Option b is the correct answer.