Respuesta :
Answer:
i = 1.13\times 10^{-4}A
Explanation:
r = radius of the coil = 4 cm = 0.04 m
Area of coil is given as
A = πr²
A = (3.14) (0.04)² = 0.005024 m²
N = Number of turns = 500
R = Resistance = 600 Ω
B = magnetic field = (0.0120)t + (3 x 10⁻⁵) t⁴
Taking derivative both side relative to "t"
[tex]\frac{dB}{dt}= (0.0120 + (12\times 10^{-5})t^{3})[/tex]
Induced current is given as
[tex]i = \left ( \frac{NA}{R} \right )\left ( \frac{dB}{dt} \right )[/tex]
[tex]i = \left ( \frac{NA}{R} \right ) (0.0120 + (12\times 10^{-5})t^{3})[/tex]
inserting the values at t = 5
[tex]i = \left ( \frac{(500)(0.005024)}{600} \right ) (0.0120 + (12\times 10^{-5})5^{3})[/tex]
[tex]i = 1.13\times 10^{-4}A[/tex]
Answer:
The current in the resistor is 1.1304 mA.
Explanation:
Given that,
Radius = 4.00 cm
Number of turns = 500
Magnetic field [tex]B=(0.0120)t+(3.00\times10^{-5})t^4[/tex]
Resistance [tex]R = 600\Omega[/tex]
Time t = 5.00 s
We need to calculate the emf
Using formula of emf
[tex]\epsilon=NA\dfrac{dB}{dt}[/tex]
Where, A = area
N = number of turns
B = magnetic field
Put the value into the formula
[tex]\epsilon = 500\times\pi\times(4.00\times10^{-2})^2\dfrac{d}{dt}((0.0120)t+(3.00\times10^{-5})t^4)[/tex]
[tex]\epsilon=500\times\pi\times(4.00\times10^{-2})^2\times(0.0120+1.2\times10^{-4})t^3[/tex]
We need to calculate the current in the resistor at t = 5.00
Using formula of current
[tex]\epsilon=IR[/tex]
[tex]I=\dfrac{\epsilon}{R}[/tex]
Where, R = resistance
Put the value into the formula
[tex]I=\dfrac{500\times\pi\times(4.00\times10^{-2})^2\times(0.0120+1.2\times10^{-4})t^3}{600}[/tex]
Put the value of t in equation (II)
[tex]I=\dfrac{500\times\pi\times(4.00\times10^{-2})^2\times(0.0120+1.2\times10^{-4})(5.00)^3}{600}[/tex]
[tex]I=0.00011304\ A[/tex]
[tex]I=1.1304\times10^{-3}\ A[/tex]
[tex]I=1.1304\ mA[/tex]
Hence, The current in the resistor is 1.1304 mA.
