Question

A coil 4.00 cm in radius, containing 500 turns, is placed in a uniform magnetic field that varies with time according to B = (0.0120 T/s)t + (3.00 x 10-5 T/s4)t4. The coil is connected to a 600 Ω resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil. What is the current in the resistor at time t = 5.00 s?

Answer 1

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"

$\frac{dB}{dt}= (0.0120 + (12\times 10^{-5})t^{3})$

Induced current is given as

$i = \left ( \frac{NA}{R} \right )\left ( \frac{dB}{dt} \right )$

$i = \left ( \frac{NA}{R} \right ) (0.0120 + (12\times 10^{-5})t^{3})$

inserting the values at t = 5

$i = \left ( \frac{(500)(0.005024)}{600} \right ) (0.0120 + (12\times 10^{-5})5^{3})$

$i = 1.13\times 10^{-4}A$

Answer 2

Answer:

The current in the resistor is 1.1304 mA.

Explanation:

Given that,

Radius = 4.00 cm

Number of turns = 500

Magnetic field $B=(0.0120)t+(3.00\times10^{-5})t^4$

Resistance $R = 600\Omega$

Time t = 5.00 s

We need to calculate the emf

Using formula of emf

$\epsilon=NA\dfrac{dB}{dt}$

Where, A = area

N = number of turns

B = magnetic field

Put the value into the formula

$\epsilon = 500\times\pi\times(4.00\times10^{-2})^2\dfrac{d}{dt}((0.0120)t+(3.00\times10^{-5})t^4)$

$\epsilon=500\times\pi\times(4.00\times10^{-2})^2\times(0.0120+1.2\times10^{-4})t^3$

We need to calculate the current in the resistor at t = 5.00

Using formula of current

$\epsilon=IR$

$I=\dfrac{\epsilon}{R}$

Where, R = resistance

Put the value into the formula

$I=\dfrac{500\times\pi\times(4.00\times10^{-2})^2\times(0.0120+1.2\times10^{-4})t^3}{600}$

Put the value of t in equation (II)

$I=\dfrac{500\times\pi\times(4.00\times10^{-2})^2\times(0.0120+1.2\times10^{-4})(5.00)^3}{600}$

$I=0.00011304\ A$

$I=1.1304\times10^{-3}\ A$

$I=1.1304\ mA$

Hence, The current in the resistor is 1.1304 mA.