A loop ABCDA, carrying current = 12A, is placed in a plane, consists of two semi-circular segments of radius and . The magnitude of the resultant magnetic field at center O is . The value of k is _______ (Given ) [2025]

Question

A loop ABCDA, carrying current $I$ = 12A, is placed in a plane, consists of two semi-circular segments of radius $R_{1} = 6 π m$ and $R_{2} = 4 π m$ . The magnitude of the resultant magnetic field at center O is $k \times 10^{- 7} T$ . The value of k is _______ (Given $μ_{0} = 4 π \times 10^{- 7} T {mA}^{- 1}$ ) [2025]

Smart Achievers · Accepted Answer

(1)

Magnetic field due to AB and CD = 0

$B = \frac{μ_{0} I}{2 r} (\frac{θ}{2 π})$ for an arc

For semi-circule $B = \frac{μ_{0} I}{4 r}$

$B_{net} = \frac{μ_{0} I}{4 {4 π}} - \frac{μ_{0} I}{4 {6 π}}$

$= \frac{μ_{0} I}{4 π} {\frac{1}{4} - \frac{1}{6}}$

$= 10^{- 7} \times 12 \times \frac{2}{24} = 10^{- 7} T \Rightarrow k = 1$

Solution

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Solution

Q. A loop ABCDA, carrying current I = 12A, is placed in a plane, consists of two semi-circular segments of radius R1=6 πm and R2=4 πm. The magnitude of the resultant magnetic field at center O is k×10–7 T. The value of k is _______ (Given μ0=4π×10–7 T mA–1) [2025]

Ans. (1) Magnetic field due to AB and CD = 0 B=μ0I2r(θ2π) for an arc For semi-circule B=μ0I4r Bnet=μ0I4{4π}–μ0I4{6π} =μ0I4π{14–16} =10–7×12×224=10–7 T ⇒ k=1

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