Question

The next two questions refer to this situation: A rectangular loop with sides of length a= 1.00 cm and b= 2.70 cm is placed near a wire that carries a current that varies as a function of time:i(t)=3.62+1.49t2where the current is in Amperes and the time is in seconds. The distance from the straight wire to the closest side of the loop is d= 0.460 centimeters.

Answer 1

Answer:

a) $1.007 * 10^{-7}$ V.m , into the page

b) $-5.39 * 10^{-8}$ V, counterclockwise

Explanation:

a) $d\Phi = B.dA\\\\B_r = \frac{\mu_0I}{2\pi r}\\\\dA = b.dr\\\\\Phi = \int\limits^b_a {B.dA} = \int\limits^{a+d}_d {\frac{\mu_0I}{2\pi r}.b.dr} =\frac{\mu_0I}{2\pi }.b.ln(\frac{a+d}{d} )$

At t = 2.90, I = 3.62 + 1.49 * (2.90)^2 = 16.15 A

$\Phi = \frac{4\pi * 10^{-7}*16.15*0.027}{2\pi} ln(1.46/0.46) = 1.007 * 10^{-7}$ V.m

Direction is into the page.

b) $emf = -d\Phi/dt = -\frac{\mu_0}{2\pi }.b.ln(\frac{a+d}{d}).(2.98t)= \\=-\frac{4\pi * 10^{-7}*0.027}{2\pi} ln(1.46/0.46)*2.98*2.90= -5.39 * 10^{-8} V$

Direction is counterclockwise.