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When a wire 1.5 m long carries a 24-A current in the +x direction in a uniform external magnetic field, the magnetic force exerted by the field on the wire is given by FBx = 0, FBy = 3.0 N, FBz = 2.0 N. When the wire is rotated until the charge carriers travel in the +y direction, the magnetic force exerted by the field on the wire is given by FBx = -3.0 N, FBy = 0, FBz = -2.0 N. Part A: Determine the components of the magnetic field.Part B: Determine the magnitude of the magnetic field.

Respuesta :

Answer:

Explanation:

Let the magnetic field be

B = [tex]B_xi +B_yj +B_z k[/tex]

For magnetic force , the expression is

F = L ( I x B )

= 1.5 ( 24i x [tex]B_xi +B_yj +B_z k[/tex] )

F = 36 B_y k - 36 B_z j

Given

F = 3 j + 2 k

Equating equal terms

we have

B_y = 2 / 36 , B_z = - 3 / 36

Now the direction of current is changed to y direction

so F = [tex]1.5[ 24j \times ( B_xi +B_yj +B_z k )][/tex]

Given

F = - 3 i - 2 k

Equating equal terms

we have

B_x = 2 / 36 , B_z = - 3 / 36

So B = 2/36 i + 2/36 j - 3/ 36 k

Magnitude of B

= 4.1 / 36 T

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