Question

Question 14 (7 points) A straight wire of length L carries a current I in the positive z-direction in a region where the magnetic field is uniform and specified by B = 3Bữ – 2Bỳ + Bĩ and B is a constant. What is the magnitude of the magnetic force on the wire? 1.0 ILB 03. 2 ILB O 3.6 ILB 4.2 ILB 0 5.0 ILB

Answer 1

Force on a current carrying wire in a magnetic field is given by

$\vec F = I(\vec L \times \vec B)$

where L is the length of the wire with direction along the direction of current.

Thus,

$\vec F = I\left [ L \hat z \times \left (3B \hat x -2B \hat y +B\hat z \right ) \right ]$

$\vec F = I\left [ 3LB \hat y +2LB \hat x \right ]$

because   $\hat z \times \hat x = \hat y$    , $\hat z \times \hat y = -\hat x$ and $\hat x \times \hat x = 0$

Thus,

$\left |\vec F \right | =\sqrt{3^{2}+2^{2}} ILB = \sqrt{13}ILB =3.6ILB$

So, option C is correct.