In the given circuit ABCD is a square of side ' a '. If magnetic induction at point O is 2 π a N μ 0 I ( 2 − 1 ) then find the value of N . Dotted line represents that wire goes upto infinity.

B due to square portion is zero B due to remaining portion is = 6 4 π ( 2 a ) μ 0 I [ sin 9 0 ∘ + sin 4 5 ∘ ] = 2 π a 3 μ 0 I ( 2 − 1 ) .

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PhysicsMagnetismBiot-savart's & Ampere's Circuital lawMedium2 minQB

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In the given circuit ABCD is a square of side ' a '. If magnetic induction at point O is $\frac{N μ _{0} I ( 2 - 1 )}{2 π a}$ then find the value of N . Dotted line represents that wire goes upto infinity.

Answer:

Solution:

B due to square portion is zero B due to remaining portion is $= 6 \frac{μ _{0} I}{4 π ( \frac{a}{2} )} [sin 9 0^{\circ} + sin 4 5^{\circ}] = \frac{3 μ _{0} I ( 2 - 1 )}{2 π a}$ .

Stream:JEESubject:PhysicsTopic:MagnetismSubtopic:Biot-savart's & Ampere's Circuital law

⏱ 2mℹ️ Source: QB

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