Infinite number of straight wires each carrying current I are equally placed as shown in the figure. Adjacent wires have current in opposite direction. Net magnetic field at point P is a)b)c)d)ZeroCorrect answer is option 'B'. Can you explain this answer?

Question

Infinite number of straight wires each carrying current I are equally placed as shown in the figure. Adjacent wires have current in opposite direction. Net magnetic field at point P is                                 a)b)c)d)ZeroCorrect answer is option 'B'. Can you explain this answer?

Answer

Magnetic field at P is equal to the magnetic field due to AB,CD,EF and so on.
Magnetic field due to AB is +Z direction.
and CD is negative −Z direction and EP is positive +Zdrection and so on.
B_P= (μ₀I/4πd₁) [sin30^°−sin(−30^°)k]+ (μ₀I/4πd₂)[sin30°−sin(−30°)l−k]+ μ0I /4πd₃ [sin30°−sin(−30°)k]+...∞
B_P=2sin30°× (μ₀I/4π) [(1/d₁)−(1/d₂)+(1/d₃)−(1/d₄)+(1/d₅)....∞]k
= μ0I/4π [(1/acos30°)−(1/2acos30°)+(1/3acos30°)−....∞]k
= μ₀I/4π(acos30°) [1−(1/2)+(1/3)−(1/4)+(1/5)....∞]k
=(2μ₀I/4π√3a) ln(1+1)k= (μ₀I/4π√3a) ln2²k=(μ₀I/4π√3a) ln4k

Answer

IT is not so hard just quite easy

https://edurev.in/studytube/it-is-not-so-hard-just-quite-easy/3e9a2db6-4c97-466f-968e-668f7ae38907_t?courseid=-1

Answer

the magnetic field at point P from the wires are given as

Therefore the magnetic field at point P is equal to

Answer

Magnetic field at P is equal to the magnetic field due to AB,CD,EF and so on.
Magnetic field due to AB is +Z direction.
and CD is negative −Z direction and EP is positive +Zdrection and so on.
B_P= (μ₀I/4πd₁) [sin30^°−sin(−30^°)k]+ (μ₀I/4πd₂)[sin30°−sin(−30°)l−k]+ μ0I /4πd₃ [sin30°−sin(−30°)k]+...∞
B_P=2sin30°× (μ₀I/4π) [(1/d₁)−(1/d₂)+(1/d₃)−(1/d₄)+(1/d₅)....∞]k
= μ0I/4π [(1/acos30°)−(1/2acos30°)+(1/3acos30°)−....∞]k
= μ₀I/4π(acos30°) [1−(1/2)+(1/3)−(1/4)+(1/5)....∞]k
=(2μ₀I/4π√3a) ln(1+1)k= (μ₀I/4π√3a) ln2²k=(μ₀I/4π√3a) ln4k

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