Respuesta :
Answer:
Explanation:
vertical magnetic field B_v = 4 x 10⁻⁵ T.
Magnetic field due to horizontal current at point 20 cm above
= (μ₀/4π ) x (2i / R)
= 10⁻⁷ x 2 x 20/ 20 x 10⁻²
= 2 x 10⁻⁵ T
It will act coming out of paper. Hence it will be normal to magnetic field given .
So resultant magnetic field
= √ (4² + 2²) x 10⁻⁵ T
= 4.47 X 10⁻⁵ T .
The resultant magnitude of the magnetic field 20 cm above the wire is; B = 44.72 μT
We are given;
Vertical magnetic field; B_v = 4 × 10^(-5) T
Horizontal Current; I = 20A
Distance; r = 20 cm = 0.2 m
Formula for the horizontal magnetic field from bio savart's law is;
B_h = (μ₀ × 2I)/(4πr)
Where;
μ₀ = 4π x 10⁻⁷ m/A.
Thus;
B_h = (4π x 10⁻⁷ × 2 × 20)/(4π × 0.2)
B_h = 2 × 10^(-5) T
Thus, resultant magnitude of magnetic field is;
B = √(2 × 10^(-5))² + (4 × 10^(-5))²
B = 44.72 × 10^(-6) T
B = 44.72 μT
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