Answer:
a)Current will flow perpendicularly.
b)Magnitude of flux will be 2.987 N m2 C−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 ; [tex]\phi = \frac{\mu_o Q{at}}{2 \pi}In (\frac{b+a}{b})[/tex]
c) The amount of the induced current that flows in the loop = [tex]3.50*10^{-8} \ \ A[/tex]
Explanation:
The magnitude of the flux through the loop is given by the equation ; [tex]\phi = \frac{\mu_o Q{at}}{2 \pi}In (\frac{b+a}{b})[/tex]
The amount of the induced current that flows in the loop can be calculated as:
Emf induced [tex]E = - \frac{d \phi}{dt}[/tex]
[tex]E = \frac{- \mu_o Q_a }{2 \pi} In (\frac{b+a}{b}) \frac{dt}{dt}[/tex]
[tex]E = \frac{- \mu_o Q_a }{2 \pi} In (\frac{b+a}{b})[/tex]
[tex]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}})[/tex]
[tex]E = -5.37*10^{-8}*1.627\\\\E = -8.737*10^{-8}\ \ V[/tex]
The current induced [tex]I = \frac{E}{R}[/tex]
[tex]\frac{-8.737*10^{-8}}{2.5} \\\\= 3.50*10^{-8} \ \ A[/tex]
