A5 cm length of wire carries a current of 3 a in the positive z direction. the force on this wire due to a magnetic field vector is vector = (-0.2 ihatbold + 0.2 jhatbold) n. if this wire is rotated so that the current flows in the positive x direction, the force on the wire is vector = 0.2 khatbold n. find the magnetic field vector.

B = 4/3 ( i + j + k )

Explanation:

I = 3 k

L = 5 x 10⁻² m

F = - 0.2 i + 0.2 j

Let magnetic field be B

B = B₁ i + B₂ j + B₃ k

F = L ( I x B )

= 5 x 10⁻² 3 k x ( B₁ i + B₂ j + B₃ k )

=   15 x 10⁻² B₁ j  -   15 x 10⁻² B₂ i

Given

F = - 0.2 i + 0.2 j

equating equal terms

15 x 10⁻² B₂ = .2

15 x 10⁻² B₁ =  .2

B₁ = .02 / 15 x 10⁻² = 20 / 15 N

B₂ = .02 / 15 x 10⁻² = 20 / 15 N

Now wire is rotated so that current flows in positive x direction

I = 3 i

F = . 2 k

F = L ( I x B )

= 5 x 10⁻² x 3 i  x ( B₁ i + B₂ j + B₃ k )

= 15 x 10⁻² B₃ k

15 x 10⁻² B₃ = .2

B₃ = .2 /15 x 10⁻²

B₃ = 20 /15

B = B₁ i + B₂ j + B₃ k

20 /15 9 ( i + j + k )

B = 4/3 ( i + j + k )

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.