Question

A flat loop of wire consisting of a single turn of cross-sectional area 7.10 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.50 T in 1.03 s. What is the resulting induced current if the loop has a resistance of 1.60

Answer 1

Answer:

The resulting induced current if the loop has a resistance of 1.60 is 0.43 mA

Explanation:

First, we will determine the induced emf. The emf is given by

$Emf = - N\frac{\Delta\phi}{\Delta t}$

Where $N$ is the number of turns

$\Delta\phi$ is the change in magnetic flux

and $\Delta t$ is the change in time

The change in magnetic flux is given by

$\Delta\phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

The magnetic flux is given by

$\phi = BA$

Where $B$ is the magnitude of magnetic field

and $A$ is the area

Therefore,

$\Delta\phi = \phi _{f} - \phi _{i} = B _{f}A - B _{i} A$

$\Delta\phi = (B _{f} - B _{i})A$

From the question

$B _{f} = 1.50 T$

$B _{i} = 0.500 T$

$A = 7.10 cm^{2} = 0.00071 m^{2}$

∴ $\Delta\phi = (1.5 - 0.5)(0.00071)$

$\Delta\phi = 0.00071 Wb$

From the question

$N = 1$

$\Delta t= 1.03 s$

Hence,

$Emf = - N\frac{\Delta\phi}{\Delta t}$ becomes

$Emf = - (1)\frac{0.00071}{1.03}$

$Emf = 0.000689 V$

Now, to determine the induced current

From Ohm's law

$Emf = IR$

Where $I$ is the current

and $R$ is the resistance

∴$I =\frac{Emf}{R}$

From the question

$R = 1.60$

Hence,

$I =\frac{0.000689}{1.60}$

$I = 0.00043 A$

$I = 0.43 mA$

Hence, the resulting induced current if the loop has a resistance of 1.60 is 0.43 mA.