A wire of arbitrary shape, which is confined to the x-y plane, carries a current i from point a to point b in the plane. show that if a uniform magnetic field b→ perpendicular to the x-y plane is present, the force that the wire experiences is the same as that which would be felt by a wire running straight from a to

b.

See explanation

Explanation:

Solution:-

- A wire of arbitrary shape,which is confined to the x-y plane,carries a current I from point A to point B in the x-y plane.

- See diagram (attached) for clarity.

- Let’s assume that the horizontal distance between A and B is "s" and the vertical distance between A and B is "d". Then for the straight line path vector ( L ):

                    L = s i^ + d j^

- The force on the straight wire with current I is then:

                    F = I * ( L x B )

Where,  L: The path vector between points A and B

             B: The magnetic field strength vector

For the curved wire vector "ds = dx i^ + dy j^" and the force on the wire is:

                   F = ∫ [ I (ds x B) = I ∫ (dx i^ + dy j^) x B

When current "I" and magnetic field "B" are uniform then we can pull both of them out of the integral. Separate the integral and calculate each differential separately:

                  F = I ∫ (dx i^) x B + I ∫ (dy j^) x B

                     = I (s i^ x B) + I ( d j^ x B ) = I ( L x B )

- The force of curved and straight line have the same force:

                 F = I ( L x B ) acting on them.

f=ma newton's second law 7x8=56n