Question

A square loop of side length a =4.5 cm is placed a distance b = 1.1 cm from a long wire carrying a current that varies with time at a constant rate, i.e. I(t) = Qt, where Q = 5.9 A/s is a constant. In what direction does this current flow?
Part (b) What is the magnitude of the flux through the loop? Select the correct expression Correct Part (c) If the loop has a resistance of 2.5 Ω, how much induced current flows in the loop

Answer 1

Answer:

a) The current will flow in a clockwise direction

b) The magnitude of the flux through the loop is given by the equation ; $\phi = \frac{\mu_o Q{at}}{2 \pi}In (\frac{b+a}{b})$

c) The amount of the induced current that flows in the loop = $3.50*10^{-8} \ \ A$

Explanation:

The magnitude of the flux through the loop is given by the equation ; $\phi = \frac{\mu_o Q{at}}{2 \pi}In (\frac{b+a}{b})$

The amount of the induced current that flows in the loop can be calculated as:

Emf induced  $E = - \frac{d \phi}{dt}$

$E = \frac{- \mu_o Q_a }{2 \pi} In (\frac{b+a}{b}) \frac{dt}{dt}$

$E = \frac{- \mu_o Q_a }{2 \pi} In (\frac{b+a}{b})$

$E = \frac{- 4 \pi *10^{-7}*5.9*4.5*10^{-2} }{2 \pi} In (\frac{(4.5+1.1)*10^{-2}}{1.1*10^{-2}})$

$E = -5.37*10^{-8}*1.627\\\\E = -8.737*10^{-8}\ \ V$

The current induced $I = \frac{E}{R}$

$\frac{-8.737*10^{-8}}{2.5} \\\\= 3.50*10^{-8} \ \ A$

Answer 2

Answer:

a)Current will flow perpendicularly.

b)Magnitude of flux will be 2.987 N m2 C−1