Question

In the opposite figure, the total magnetic flux density at point (O) equals: Knowing that \( \left(\mu_{\text {air }}=4 \pi \times 10^{-7} \mathrm{~Wb} /\right. \) Amp. \( \mathrm{m}, \pi=3.14 \) ) (A) \( 1.442 \times 10^{-5} \mathrm{~T} \) (B) \( 5 \times 10^{-6} \mathrm{~T} \) (C) \( 9.42 \times 10^{-5} \mathrm{~T} \) (D) \( 4.42 \times 10^{-6} \mathrm{~T} \)

          In the opposite figure, the total magnetic flux density at point (O) equals:
Knowing that \( \left(\mu_{\text {air }}=4 \pi \times 10^{-7} \mathrm{~Wb} /\right. \) Amp. \( \mathrm{m}, \pi=3.14 \) )
(A) \( 1.442 \times 10^{-5} \mathrm{~T} \)
(B) \( 5 \times 10^{-6} \mathrm{~T} \)
(C) \( 9.42 \times 10^{-5} \mathrm{~T} \)
(D) \( 4.42 \times 10^{-6} \mathrm{~T} \)

In the opposite figure, the total magnetic flux density at point (O) equals:
Knowing that (μair =4 π× 10^-7 Wb /. Amp. m, π=3.14 )
(A) 1.442 × 10^-5 T
(B) 5 × 10^-6 T
(C) 9.42 × 10^-5 T
(D) 4.42 × 10^-6 T

Added by Gregg M.

Essential College Physics

Andrew F. Rex, Richard Wolfson 1st Edition

Chapter 19

Instant Answer

Solved by Expert Christian Dell

Step 1

There are two current-carrying wires: one straight wire with current \( I_1 = 5 \, \text{A} \) and one semicircular wire with current \( I_2 = 12 \, \text{A} \). Show more…

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In the opposite figure, the total magnetic flux density at point (O) equals: Knowing that \( \left(\mu_{\text {air }}=4 \pi \times 10^{-7} \mathrm{~Wb} /\right. \) Amp. \( \mathrm{m}, \pi=3.14 \) ) (A) \( 1.442 \times 10^{-5} \mathrm{~T} \) (B) \( 5 \times 10^{-6} \mathrm{~T} \) (C) \( 9.42 \times 10^{-5} \mathrm{~T} \) (D) \( 4.42 \times 10^{-6} \mathrm{~T} \)