The magnetic field B due to a current-carrying circular loop of the radius 12 cm at its centre is 0.5 × 10 – 4 T . The magnetic field due to this loop at a point on the axis at a distance of 5 cm from the centre -

B 0 = μ 0 I 2 a At axial point B = μ 0 I a 2 2 ( a 2 + x 2 ) 3 / 2 B B 0 = a 3 ( a 2 + x 2 ) 3 / 2 ⇒ B = B 0 a 3 ( a 2 + x 2 ) 3 / 2 = 0.5 × 10 – 4 × ( 12 c m ) 3 ( 144 c m 2 + 25 c m 2 ) 3 / 2 = 3.9 × 10 – 5 T .

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Physics - Magnetism Question with Solution | TestHub

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

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The magnetic field $B$ due to a current-carrying circular loop of the radius $12 cm$ at its centre is $0.5 \times 10^{- 4} T$ . The magnetic field due to this loop at a point on the axis at a distance of $5 cm$ from the centre -

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Answer:

Solution:

$B_{0} = \frac{μ_{0} I}{2 a}$

At axial point

$B = \frac{μ_{0} I a^{2}}{2 (a^{2} + x^{2})^{3 / 2}}$

$\frac{B}{B_{0}} = \frac{a^{3}}{(a^{2} + x^{2})^{3 / 2}}$

$\Rightarrow B = B_{0} \frac{a^{3}}{(a^{2} + x^{2})^{3 / 2}}$

$= 0.5 \times 10^{- 4} \times \frac{(12 c m)^{3}}{(144 c m^{2} + 25 c m^{2})^{3 / 2}}$

$= 3.9 \times 10^{- 5} T .$

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

⏱ 2mℹ️ Source: PYQ_2020

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