A circular conductor of radius ro- 1 cm has an internal field-H - 104ro 1a2sin aro - roacos aro a- A-mwhere- a - -2r0

According to Ampere’s law, I_enc = ∮ r= r_0 H .dl r_o = 10^ 2 m ∫_0^2π10^4r_0 ( 4r^2_0π ^2sinπ2 2r^2_0πcosπ2 ) . r_0 dϕ = 10^4 ∫_0^2π4r^2_0π ^2 d ϕ l_enc = ...

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Application of Ampere's Law : Magnetic Field Due to Circular Conductor

A circular co...

Question

A circular conductor of radius $r_{o} = 1 c m$ has an internal field,

$¯ ¯¯¯ ¯ H = \frac{10^{4}}{r_{o}} (\frac{1}{a^{2}} sin a r_{o} - \frac{r_{o}}{a} cos a r_{o}) {ˆ a}_{ϕ} (A / m)$

where, $a = \frac{π}{2 r_{0}}$

8 π A m p e r e

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\frac{8}{π} A m p e r e

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4 π A m p e r e

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\frac{4}{π} A m p e r e

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Solution

The correct option is B $\frac{8}{π} A m p e r e$
According to Ampere’s law,

$I_{e n c} = \oint r = r_{0} \to H . d l$
$r_{o} = 10^{- 2} m$

$\int_{0}^{2 π} \frac{10^{4}}{r_{0}} (\frac{4 r_{0}^{2}}{π^{2}} sin \frac{π}{2} - \frac{2 r_{0}^{2}}{π} cos \frac{π}{2}) . r_{0} d ϕ$

$= 10^{4} \int_{0}^{2 π} \frac{4 r_{0}^{2}}{π^{2}} d ϕ$

$l_{e n c} = 10^{4} . \frac{4 r_{0}^{2}}{π} \times 2 = \frac{8}{π} A m p e r e$

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A circular conductor of radius ro=1cm has an internal field, ¯¯¯¯¯H=104ro(1a2sinaro−roacosaro)ˆaϕ(A/m) where, a=π2r0

The correct option is B 8πAmpereAccording to Ampere’s law, Ienc=∮r=r0→H.dl ro=10−2 m ∫2π0104r0(4r20π2sinπ2−2r20πcosπ2).r0dϕ =104∫2π04r20π2dϕ lenc=104.4r20π×2=8πAmpere

A circular conductor of radius $r_{o} = 1 c m$ has an internal field,

$¯ ¯¯¯ ¯ H = \frac{10^{4}}{r_{o}} (\frac{1}{a^{2}} sin a r_{o} - \frac{r_{o}}{a} cos a r_{o}) {ˆ a}_{ϕ} (A / m)$

where, $a = \frac{π}{2 r_{0}}$