A conductor carrying current 60 A is in the form of a semicircle AB of radius R and lying in xy-plane

Question

A conductor carrying current 60 A is in the form of a semicircle AB of radius R and lying in xy-plane with its centre 'O' at origin as shown in the figure. The magnitude of $\oint \vec{B} \cdot d \vec{ℓ}$ for the circle $x^{2} + z^{2} = 3 R^{2}$ in $x z$ -plane due to current in curve AB is 10 $n μ_{0}$ . Find the value of $n$ ( $n$ is an integer)

Smart Achievers · Accepted Answer

(3)

$\oint {\vec{B}}_{net} \cdot d \vec{l} = \oint {\vec{B}}_{1} \cdot d \vec{l} + \oint {\vec{B}}_{2} \cdot d \vec{l}$

${\vec{B}}_{1}$ is magnetic field due to straight part,

${\vec{B}}_{2}$ is magnetic field due to curved part

$\oint {\vec{B}}_{net} \cdot d \vec{l} = - B_{1} (2 π R \sqrt{3}) + \oint {\vec{B}}_{2} \cdot d \vec{l}$

$= - \frac{μ_{0} I}{4 π R \sqrt{3}} (2 π R \sqrt{3}) + \oint {\vec{B}}_{2} \cdot d \vec{l}$

$0 = - \frac{μ_{0} I}{2} + \oint {\vec{B}}_{2} \cdot d \vec{l}$

Solution

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Solution

Ans. (3) ∮B→net·dl→=∮B→1·dl→+∮B→2·dl→ B→1 is magnetic field due to straight part, B→2 is magnetic field due to curved part ∮B→net·dl→=-B1(2πR3)+∮B→2·dl→ =-μ0I4πR3 (2πR3)+∮B→2·dl→ 0=-μ0I2+∮B→2·dl→

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