Respuesta :
Answer:
The magnitude of the emf induced in the loop is 2.4 × 10⁻⁵ V
Explanation:
The magnitude of the induced emf of a loop moving relative to a wire is given as follows;
The given parameters are;
I = 50 A, a = 50 cm, and b = 6.0 cm
The constant rate of change of x, v = 20 cm/s
Motional emf formula
The magnetic field due to the straight wire, B = μ₀·i/(2·π·x)
The flux given by the rectangular loop, Ф = BA = μ₀·i/(2·π·x) × a × b
Therefore, we get;
[tex]e = \dfrac{-d\phi}{dt} = \dfrac{-\mu_0 \cdot i \cdot a\cdot b}{2 \cdot \pi} \times \dfrac{d}{dt} \left (\dfrac{1}{x} \right ) = \dfrac{-\mu_0 \cdot i \cdot a\cdot b}{2 \cdot \pi} \times \left (\dfrac{1}{x^2} \right ) \dfrac{dx}{dt}[/tex]
[tex]\because \dfrac{dx}{dt} = v[/tex]
Therefore;
[tex]e = \dfrac{-\mu_0 \cdot i \cdot a\cdot b}{2 \cdot \pi} \times \left (\dfrac{1}{x^2} \right ) \dfrac{dx}{dt} = \dfrac{-\mu_0 \cdot i \cdot a\cdot b}{2 \cdot \pi} \times \left (\dfrac{1}{x^2} \right ) \times v \left(\because \dfrac{dx}{dt} = v \right)[/tex]
μ₀ = 4·π × 10⁻⁷ H/m
We get;
[tex]e = \dfrac{4\cdot \pi \times 10^{-7} \, H/m\times50 \, A \times 0.5 \, m\times 0.06 \, m}{2 \cdot \pi} \times \left (\dfrac{1}{(0.05 \, m)^2} \right ) \times 0.2 \ m/s = 2.4 \times 10^{-5} \, V[/tex]
The induced emf, e = 2.4 × 10⁻⁵ V
