Answer :
Answer:
Therefore,
The magnitude of the force per unit length that one wire exerts on the other is
[tex]\dfrac{F}{l}=2.79\times 10^{-5}\ N/m[/tex]
Explanation:
Given:
Two long, parallel wires separated by a distance,
d = 3.50 cm = 0.035 meter
Currents,
[tex]I_{1}=1.55\ A\\I_{2}=3.15\ A[/tex]
To Find:
Magnitude of the force per unit length that one wire exerts on the other,
[tex]\dfrac{F}{l}=?[/tex]
Solution:
Magnitude of the force per unit length on each of @ parallel wires seperated by the distance d and carrying currents I₁ and I₂ is given by,
[tex]\dfrac{F}{l}=\dfrac{\mu_{0}\times I_{1}\times I_{2}}{2\pi\times d}[/tex]
where,
[tex]\mu_{0}=permeability\ of\ free\ space =4\pi\times 10^{-7}[/tex]
Substituting the values we get
[tex]\dfrac{F}{l}=\dfrac{4\pi\times 10^{-7}\times 1.55\times 3.15}{2\pi\times 0.035}[/tex]
[tex]\dfrac{F}{l}=2.79\times 10^{-5}\ N/m[/tex]
Therefore,
The magnitude of the force per unit length that one wire exerts on the other is
[tex]\dfrac{F}{l}=2.79\times 10^{-5}\ N/m[/tex]
Answer:
The force per unit length is [tex]2.79 \times 10^{-5}~N[/tex].
Explanation:
As depicted in the figure, the current ([tex]I_{1}[/tex]) through wire 1 is 1.55 A upwards and the current ([tex]I_{2}[/tex]) through wire 2 is 3.15 A downwards. The wires are 3.50 cm apart.
From Biot-Savart's law, we know that the magnetic field (B) due to a current carrying conductor carrying a current I at a distance r from the conductor is given by
[tex]\vec{B} = \dfrac{\mu_{0}I}{4 \pi} \oint \dfrac{\vec{dl} \times \hat{r}}{r^{2}}[/tex]
According to the figure, the magnetic field (B) at 3.5 cm right to wire 1 due to it can be written as
[tex]B = \dfrac{\mu_{0}I_{1}}{2\pi \times 0.035~m} = \dfrac{4 \pi \times10^{-7} N~A^{-2}\times 1.55~A}{2 \pi \times 0.035~m} = 88.57 \times 10^{-7} T[/tex]
andthe direction will be perpendicular to the screen and towards the screen (as depicted by the [tex]\bigotimes[/tex] symbol in the figure).
Also we know that the force 'F' experience by a current carrying conductor of length 'L' having current 'I' and placed in a magnetic field 'B' is given by
[tex]F = I~L~B[/tex]
So the force per unit length ([tex]F_{21}[/tex]) experienced by wire 2 due to the magnetic field due to wire 1 is given by
[tex]F_{21} = \dfrac{F}{L} = I_{2} \times B = 3.15~A \times 88.57 \times 10^{-7}~T = 2.79 \times 10^{-5} N[/tex]
and the direction of the force will be right to wire 2.
