Question

A 143 ‑turn circular coil of radius 3.43 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicular to the plane of the coil. The coil is connected to a 10.7 Ω resistor to create a closed circuit. During a time interval of 0.115 s, the magnetic field strength decreases uniformly from 0.523 T to zero. Find the energy, in millijoules, that is dissipated in the resistor during this time interval.

Answer 1

Answer:

61.9 mJ

Explanation:

We can start by calculating the emf induced in the coil by the changing magnetic field, which is given by:

$\epsilon = -\frac{\Delta \Phi}{\Delta t}$

where

$\Delta \Phi$ is the change in flux linkage through the coil

$\Delta t = 0.115 s$ is the time interval

The flux linkage through the coil is given by

$\Phi = NAB$

where

N = 143 is the number of turns

A is the area of the coil

B is the strength of the magnetic field

The radius of the coil is

r = 3.43 cm = 0.0343 m

So its area is

$A=\pi r^2 = \pi (0.0343)^2=3.70\cdot 10^{-3} m^2$

So, the initial flux linkage is

$\Phi_i = NAB_i = (143)(3.70\cdot 10^{-3})(0.523)=0.276 Wb$

where $B_i=0.523 T$ is the initial strength of the magnetic field.

The final strength of the magnetic field is zero:

$B_f=0$

So, the final flux linkage is zero: $\Phi_f = 0$

Therefore, the change in flux linkage through the coil is

$\Delta \Phi = \Phi_f -\Phi_i = 0-0.276=-0.276 Wb$

The flux linkage decreases uniformly, so the emf induced in the coil is constant and it is given by

$\epsilon = - \frac{\Delta \Phi}{\Delta t}=-\frac{-0.276}{0.115}=2.4 V$

This is also the potential difference across the resistor of the closed circuit connected to the coil, $V=\epsilon$. The resistance is

$R=10.7\Omega$

So, the power dissipated through the resistor is

$P=\frac{V^2}{R}=\frac{(2.4)^2}{10.7}=0.538 W$

And therefore, the energy dissipated in the time interval of 0.115 s is

$E=P\Delta t =(0.538)(0.115)=0.0619 J = 61.9 mJ$

Answer 2

The energy dissipated in the resistor is; 61.9 mJ

What is the dissipated Energy?

We are given;

Number of turns; N = 143

Resistance; R = 10.7 Ω

Radius; r = 3.43 cm = 0.0343 m

Initial magnetic field strength; B₁ = 0.532 T

Formula for initial flux linkage is;

Ф₁ = NAB₁

Where A is Area = πr² = π * 0.0343² = 3.7 × 10⁻³ m²

Thus;

Ф₁ = 143 * 3.7 × 10⁻³ * 0.532

Ф₁ = 0.276 Wb

Since the magnetic field strength decreases uniformly from 0.523 T to zero. It means that the final flux linkage is; Ф₂ = 0 Wb

Thus;

Change in flux Linkage; ΔФ = Ф₂ - Ф₁

ΔФ = 0 - 0.276

ΔФ = -0.276 Wb

The EMF induced in the coil is gotten from;

ε = -ΔФ/Δt

where Δt is the time interval which we are given as 0.115 s

ε = -(-0.276/0.115)

ε = 2.4 V

Power dissipated is;

P = (EMF)²/R

P = 2.4²/10.7

P = 0.538 W

Finally, energy dissipated is;

E = P * Δt

E = 0.538 * 0.115

E = 0.0619 J = 61.9 mJ

Read more about Energy at; https://brainly.com/question/21222010