a) The magnetic field created by a current-carrying wire is proportional to the current:
B ∝ I, B = magnetic field strength, I = current
The magnetic force acting on a current-carrying wire immersed in a magnetic field is proportional to the current and the magnetic field strength:
F ∝ IB, F = magnetic force, I = current, B = magnetic field strength
Let's focus on wire 1.
Since wire 2's current is doubled, wire 2 produces a magnetic field twice as strong as before.
Wire 1's current is also doubled, therefore we now have a wire having twice as strong a current immersed in twice as strong a magnetic field. The magnetic force on wire 1 (and you can make a similar argument for wire 2) will be four times as strong as before.
b) The general formula for the magnetic force acting on a current-carrying wire immersed in a magnetic field is given by:
F = IL×B
F = magnetic force vector
I = current
L = vector having a magnitude equal to wire length and representing direction of current
B = magnetic field vector
Note we are taking a cross product of the IL and B vectors, not the product of two scalar quantities.
The very nature of the cross product means that if L and B are parallel to each other, F = 0N
(a) The magnitude of the force that one wire exerts on the other doubles.
(b) The force on the wire will be zero.
The force between current carrying conductors is given as;
[tex]F = \frac{\mu_o I_1 I_2 l}{2\pi r}[/tex]
where;
When the length of the wire is doubled and the distance of separation remains constant, the magnitude of the force that one wire exerts on the other doubles.
F = BIlsin(θ)
at parallel field, θ = 0
F = 0
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