Question

a long straight wire is carrying a current of 25A , a rectangular coil of length 25 cm and breadth 10 cm, carrying a current of 15 A is replaced 2cm from wire , the force on the coil is
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Answer 1

Answer:

The wire carrying 25 A current produces a magnetic field in its surrounding

the wires AB and CD are equal in length and carry current in opposite direction

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Answer 2

Concept:

• Magnetic field due to a current-carrying loop

Given:

• Current in coil I₁= 25 A
• Length of coil = 25 cm = 0.25 m
• Breadth of coil = 10 cm = 0.1 m
• Current in wire I₂ =  15 A
• Distance between wire and coil d = 2 cm = 0.02 m and for other side 10+2 = 12cm = 0.12m

Find:

• Force on the coil

Solution:

F = μ I₁I₂/ 2πd

F₁ = μ *25*15/2π*0.02

F₁ = 375*10^-5 N

F₂ = μ I₁I₂/ 2πd

F₂ = μ 25*15 / 2π* 0.12

F₂ = 62.5 * 10^-5 N

Fnet = F₁ - F₂ = 312.5 * 10^-5 N

The force on the coil is 312.5 * 10^-5 N

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