The magnetic force between two wires is 0.052 N which is attract each other.
We need to know about magnetic force on a current-carrying wire formula to solve this problem. The magnetic force on two wires with same direction of current is
F = μ₀ . I1 . I2 . L / ( 2π . r )
where μ₀ is vacuum permeability (4π×10‾⁷ H/m) F is the magnetic force, I is current, L is the length of wire, r is distance of 2 wires.
From the question above, we know that:
L = 25 m
r = 6 cm
I1 = I2 = 25 A
By substituting the parameter, we get
F = μ₀ . I1 . I2 . L / ( 2π . r )
F = 4π×10‾⁷ . 25 . 25 . 25 / (2π . 0.06)
F = 0.052 N
Hence, the force between two wires is 0.052 N which is attract each other.
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Answer:15.20 m
Explanation:
Given
initial velocity 
inclined length=10 m

inclination 







So Particle launches with a speed of 5.31 m/s at an angle of \theta 



Total height raised is 
b. deflects. A DC motor's magnet has a magnetic field that <em><u>deflects</u></em> the wire loop's magnetic field, therefore causing the loop to rotate.
When a conductor, through which an electric current passes, is immersed in a magnetic field, it experiences a deflects force according to the Law of Lorentz creating as result that the loop rotate.
Answer:
Resultant velocity will be equal to 6.10 m/sec
Explanation:
We have given a motorbike is traveling with 5 m/sec in east
And a current is flowing at a rate of 3.5 m /sec in north
We know that east and north is perpendicular to each other
So resultant velocity will be vector sum of both velocity
So resultant velocity 
So resultant velocity will be equal to 6.10 m/sec