Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Solution

Given,
Current on first conductor, I₁ = 8.0 A
Current on second conductor, I₂ = 5 A
Distance between the conductors, r = 4.0 cm = $4 \times 10^{- 2}$
Length of the wire, $l = 10 cm = 0.1 m$
Force on length l is given by,

$F = \frac{μ_{0}}{2 π} \frac{I_{1} I_{2} l}{r}$
$= \frac{(4 π \times 10^{- 7}) \times 8 \times 5 \times 0.1}{2 π \times 0.04} = 2 \times 10^{- 5} N$

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M₁ and M₂ have the following particulars:
R₁ = 10 Ω N₁ = 30,
A₁ = 3.6 x 10^–3 m², B₁ = 0.25 T
R₂ = 14 Ω; N₂ = 42,
A₂ = 1.8 x 10^–3 m², B₂ = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M_1.

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Moving Charges And Magnetism

Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Some More Questions From Moving Charges And Magnetism Chapter

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