Two long parallel straight wires X and Y separated by a distance of 5 cm in air carry currents of 10 A and 5 A respectively in opposite directions. Calculate the magnitude and direction of the force on a 20 cm length of the wire Y.

Solution

Given, two current carrying parallel wires,
Distance between the wire , r = 5 cm
Current carried by first wire = 10 A
Current carried by second wire = 5A
Length of the wire, l = 20 cm

The Force between two wires can be found using the formula,

F = \frac{μ_{0} I_{1} I_{2} l}{2 πr} = \frac{4 π \times 10^{- 7} \times 10 \times 5 \times 0.20}{2 π \times 0.05} = 4 \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 parallel straight wires X and Y separated by a distance of 5 cm in air carry currents of 10 A and 5 A respectively in opposite directions. Calculate the magnitude and direction of the force on a 20 cm length of the wire Y.

Some More Questions From Moving Charges And Magnetism Chapter

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