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.

Respuesta :

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 :

[tex]B_x=0.146\ T[/tex]

[tex]B_y=0.177\ T[/tex]

[tex]B_z=0.2\ T[/tex]

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 :

[tex]F=I(L\times B)[/tex]

[tex]F=3.35(0.3k\times (0.146i+0.177j+0.2k))[/tex]

Since,

[tex]i\times j=k\\j\times k=i\\k\times i=j[/tex]

[tex]F=0.1467j-0.1778i[/tex]

[tex]|F|=\sqrt{0.1467^2+(0.1778)^2}[/tex]

|F| = 0.446 N

Direction of force:

[tex]\theta=tan^{-1}\dfrac{F_y}{F_x}[/tex]

[tex]\theta=tan^{-1}\dfrac{0.1467}{-0.1778}[/tex]

[tex]\theta=-39.52^{\circ}[/tex]

Hence, this is the required solution.