The magnetic field due to current carrying circular loop of the radius 12 c m at its centre is 0.50 × 1 0 - 4 T . The magnetic field due to this loop at a point on the axis at a distance of 5 c m from the centre is

Use B a x i s B c e n t r e = R 2 R 2 + x 2 3 / 2 B a x i s 0.5 × 10 - 4 = 12 2 12 2 + 5 2 3 / 2 B a x i s = 0.5 × 10 - 4 12 13 3 B a x i s = 3.5 × 10 - 5 T

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

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The magnetic field due to current carrying circular loop of the radius $12 c m$ at its centre is $0.50 \times 1 0^{- 4} T$ . The magnetic field due to this loop at a point on the axis at a distance of $5 c m$ from the centre is

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Solution:

Use $\frac{B_{a x i s}}{B_{c e n t r e}} = {(\frac{R^{2}}{R^{2} + x^{2}})}^{3 / 2}$

$\frac{B_{a x i s}}{0.5 \times 10^{- 4}} = {(\frac{12^{2}}{12^{2} + 5^{2}})}^{3 / 2}$

$B_{a x i s} = (0.5 \times 10^{- 4}) {(\frac{12}{13})}^{3}$

$B_{a x i s} = 3.5 \times 10^{- 5} T$

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

⏱ 2mℹ️ Source: PYQ_2020

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