Answer:
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.
