Question

Two long straight parallel wires seperated by 20cm apart carry current of 20A and 1OA in opposite directions. What is the magnitude of the force on 1m length of the 20A wire

Answer 1

Answer:

Force is 2 × 10^-4 N

Explanation:

${ \bf{F = \frac{ \gamma_{o} I _{1} I _{2} l}{2\pi r} }}$

F is force

I is current

r is separation distance

gamma is a constant

l is wire length

$F = \frac{(4\pi \times {10}^{ - 7}) \times (20) \times (10) \times 1 }{2\pi \times 0.2} \\ \\ F = 2 \times {10}^{ - 4} \: \: newtons$

Answer 2

The magnitude of the force on 1m length of the 20A wire is 2×10⁻⁴N.

To determine the answer, we need to know about magnetic field due to a current carrying wire and magnetic force.

What is the magnetic field produced due to a straight current carrying wire of current I₁?

The magnetic field due to a straight wire having current I₁ at a perpendicular distance (say) 'd'  is μ₀I₁ / 2$\pi$d.

What is the magnetic force experienced by a straight wire of current I₂?

Magnetic force on a current carrying wire of length 'L' and having current I₂ is I₂(L×B). Where B is the magnetic field at that wire.

What is the magnetic field on 20A wire due to 10A wire?

• d= 0.2m and I₁= 10A

• B= μ₀I₁ / 2$\pi$d = 4$\pi$ ×10⁻⁷× 10/ 2$\pi$×0.2 = 10⁻⁵ T

What is the force on 20A wire?

• Current (I₂) = 20A, B = 10⁻⁵T and L= 1m
• Magnitude of force on the 20 A wire of length 1m = I₂×L×B

  ( Here L and B are perpenicular to each other so L×B= L·B)    

                              = 20×1×10⁻⁵

                              =2×10⁻⁴N

Thus, we can conclude that the magnitude of the force on 1m length of the 20A wire is 2×10⁻⁴N.

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