Question

A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a constaent rate of 9 A/s. (Assume the loop and the wire lie in the same plane. Further assume I- 0at t 0. Enter the magnitudes.) (a) Use Ampère's law and find the magnetic feld from the current in the wire as a function of time and dstance from the wire. (Use the folowing as necessary: j is in meters and t is in seconds. Do not include units in your answer) , and t. Assume 8 is in teslas, r Br, t) (b) Determine the magnetic flux through the loop. (Use the following as necessary: and t. Assume is in T m and t is in seconds. Do not include units in your answer) (c) If the loop has a resistance of 3 0, how much induced current flows in the loop (in nA)) nA

Answer 1

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

$\int _{0}^{i}di =9\int _{0}^{t}dt$

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

$B = \frac{\mu_{0}i}{2\pi r}$

$B = \frac{\mu_{0}\times 9t}{2\pi r}$

(b)

Magnetic flux,

$\phi=\int B\times a dr$

$\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr$

$\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )$

$\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)$

$\phi = 1.89 \times 10^{-7}t$

(c)

R = 3 ohm

$e = -\frac{d\phi}{dt}$

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