Question

The magnetic field in a region of space is specified by its x, y, and z components which are respectively Bx = 0.146 T, By = 0.177 T, and Bz = 0.200 T. If a 30.0-cm wire carrying a current of 3.35 A is oriented along the z-axis with the current in the −z direction, determine the magnitude and direction of the force acting on the wire.

Answer 1

Explanation:

It is given that, the magnetic field in a region of space is specified by its x, y, and z components are given by :

$B_x=0.146\ T$

$B_y=0.177\ T$

$B_z=0.2\ T$

Length of the wire, L = 30 cm = 0.3 m (along z axis)

Current flowing through the wire, I = 3.35 A (along -z axis)

Magnetic force is given by :

$F=I(L\times B)$

$F=3.35(0.3k\times (0.146i+0.177j+0.2k))$

Since,

$i\times j=k\\j\times k=i\\k\times i=j$

$F=0.1467j-0.1778i$

$|F|=\sqrt{0.1467^2+(0.1778)^2}$

|F| = 0.446 N

Direction of force:

$\theta=tan^{-1}\dfrac{F_y}{F_x}$

$\theta=tan^{-1}\dfrac{0.1467}{-0.1778}$

$\theta=-39.52^{\circ}$

Hence, this is the required solution.