Answer:
The magnitude of the current is 5.45 mA.
Explanation:
Given that,
Resistance = 10.0 ohm
Radius = 0.10 m
Magnetic field = 1.0 T
Angle = 30°
Increase magnetic field = 7.0 T
Time t = 3.0 s
Number of turns = 1
We need to calculate the initial flux
Using formula of flux
[tex]\phi=NB_{1}A\cos\theta[/tex]
Put the value into the formula
[tex]\phi=1\times1.0\times\pi\times(0.10)^2\times\cos30^{\circ}[/tex]
[tex]\phi=1\times1.0\times\pi\times(0.10)^2\times\dfrac{\sqrt{3}}{2}[/tex]
[tex]\phi=0.027\ wb[/tex]
We need to calculate the final flux
[tex]\phi=1\times7.0\times\pi\times(0.10)^2\times\dfrac{\sqrt{3}}{2}[/tex]
[tex]\phi=0.1904\ wb[/tex]
We need to calculate the induced emf
Using formula of emf
[tex]\epsilon=\dfrac{\phi_{f}-\phi_{i}}{t}[/tex]
Put the value into the formula
[tex]\epsilon=\dfrac{0.1904-0.027}{3.0}[/tex]
[tex]\epsilon=0.0545\ V[/tex]
We need to calculate the current
Using formula of current
[tex]I=\dfrac{\epsilon}{R}[/tex]
Put the value into the formula
[tex]I=\dfrac{0.0545}{10.0}[/tex]
[tex]I=5.45\ mA[/tex]
Hence, The magnitude of the current is 5.45 mA.