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.

Explanation:

Let the magnetic field be

B =

For magnetic force , the expression is

F = L ( I x B )

= 1.5 ( 24i x )

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 =

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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