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

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 ((b+a)/(b))`

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

Step-by-step explanation:

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

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

Emf induced
`E = - (d \phi)/(dt)`

`E = (- \mu_o Q_a )/(2 \pi) In ((b+a)/(b)) (dt)/(dt)`

`E = (- \mu_o Q_a )/(2 \pi) In ((b+a)/(b))`

`E = (- 4 \pi *10^(-7)*5.9*4.5*10^(-2) )/(2 \pi) In (((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 = (E)/(R)`

`(-8.737*10^(-8))/(2.5) \\\\= 3.50*10^(-8) \ \ A`

Answer 2

a)Current will flow perpendicularly.

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