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

The current in the resistor is 1.1304 mA.

Step-by-step explanation:

Given that,

Radius = 4.00 cm

Number of turns = 500

Magnetic field
`B=(0.0120)t+(3.00*10^(-5))t^4`

Resistance
`R = 600\Omega`

Time t = 5.00 s

We need to calculate the emf

Using formula of emf

`\epsilon=NA(dB)/(dt)`

Where, A = area

N = number of turns

B = magnetic field

Put the value into the formula

`\epsilon = 500*\pi*(4.00*10^(-2))^2(d)/(dt)((0.0120)t+(3.00*10^(-5))t^4)`

`\epsilon=500*\pi*(4.00*10^(-2))^2*(0.0120+1.2*10^(-4))t^3`

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

Using formula of current

`\epsilon=IR`

`I=(\epsilon)/(R)`

Where, R = resistance

Put the value into the formula

`I=(500*\pi*(4.00*10^(-2))^2*(0.0120+1.2*10^(-4))t^3)/(600)`

Put the value of t in equation (II)

`I=(500*\pi*(4.00*10^(-2))^2*(0.0120+1.2*10^(-4))(5.00)^3)/(600)`

`I=0.00011304\ A`

`I=1.1304*10^(-3)\ A`

`I=1.1304\ mA`

Hence, The current in the resistor is 1.1304 mA.

Answer 2

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

Step-by-step 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"

`(dB)/(dt)= (0.0120 + (12* 10^(-5))t^(3))`

Induced current is given as

`i = \left ( (NA)/(R) \right )\left ( (dB)/(dt) \right )`

`i = \left ( (NA)/(R) \right ) (0.0120 + (12* 10^(-5))t^(3))`

inserting the values at t = 5

`i = \left ( ((500)(0.005024))/(600) \right ) (0.0120 + (12* 10^(-5))5^(3))`

`i = 1.13* 10^(-4)A`