Respuesta :
Answer:
Explanation:
side of the square loop, a = 7 cm
distance of the nearest side from long wire, r = 2 cm = 0.02 m
di/dt = 9 A/s
Integrate on both the sides
[tex]\int _{0}^{i}di =9\int _{0}^{t}dt[/tex]
i = 9t
(a) The magnetic field due to the current carrying wire at a distance r is given by
[tex]B = \frac{\mu_{0}i}{2\pi r}[/tex]
[tex]B = \frac{\mu_{0}\times 9t}{2\pi r}[/tex]
(b)
Magnetic flux,
[tex]\phi=\int B\times a dr[/tex]
[tex]\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr[/tex]
[tex]\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )[/tex]
[tex]\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)[/tex]
[tex]\phi = 1.89 \times 10^{-7}t[/tex]
(c)
R = 3 ohm
[tex]e = -\frac{d\phi}{dt}[/tex]
magnitude of voltage is
e = 1.89 x 10^-7 V
induced current, i = e / R = (1.89 x 10^-7) / 3
i = 6.3 x 10^-8 A
