A very long straight conducting wire, lying along the z -axis, carries a current of 2 A . The integral ∫ B → . d ℓ → is computed along the straight line PQ , where P has the coordinates ( 2 cm , 0 , 0 ) and Q has the coordinates ( 2 cm , 2 cm , 0 ) . The integral has the magnitude (in T m )

∫ PAQ B → . d ℓ → = ∫ RBQ B → . d ℓ → ∫ PCR B → . d ℓ → = 0 = μ 0 I 2 πR π 4 R = 4 π × 1 0 - 7 2 8 = π × 1 0 - 7 T m

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PhysicsMagnetismBiot-savart's & Ampere's Circuital lawEasy2 minPYQ_2020

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A very long straight conducting wire, lying along the $z$ -axis, carries a current of $2 A$ . The integral $\int \vec{B} . d \vec{ℓ}$ is computed along the straight line $PQ$ , where $P$ has the coordinates $(2 cm, 0, 0)$ and $Q$ has the coordinates $(2 cm, 2 cm, 0)$ . The integral has the magnitude (in $T m$ )

Options:

Answer:

Solution:

$\int_{PAQ} \vec{B} . d \vec{ℓ} = \int_{RBQ} \vec{B} . d \vec{ℓ} (\int_{PCR} \vec{B} . d \vec{ℓ} = 0)$

$= (\frac{μ_{0} I}{2 πR}) (\frac{π}{4} R)$

$= \frac{(4 π \times 1 0^{- 7}) (2)}{8} = π \times 1 0^{- 7} T m$

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

⏱ 2mℹ️ Source: PYQ_2020

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