Ask question

Ask question

All categories

3 years ago

15

One hundred turns of insulated copper wire are wrapped around an iron core of cross-sectional area 0.100m2. As the magnetic fiel

d along the coil axis changes from 0.5 T to 1.00T in 4s, the voltage induced is:

1 answer:

bagirrra123 [75]3 years ago

6 0

Answer:

The voltage induced in the coil is 1.25 V.

Explanation:

Given;

number of turns, N = 100 turns

cross sectional area of the copper coil, A = 0.1 m²

initial magnetic field, B₁ = 0.5 T

final magnetic field, B₂ = 1.00 T

duration of change in magnetic field, dt = 4 s

The induced emf in the coil is calculated as;

$emf = -N\frac{\delta \phi}{\delta t} \\\\emf = - N (\frac{\delta B}{\delta t}) A\\\\emf = -N (\frac{B_1 -B_2}{\delta t} )A\\\\emf = N(\frac{B_2-B_1}{\delta t} )A\\\\emf = 100(\frac{1-0.5}{4} )0.1\\\\emf = 1.25 \ Volts$

Therefore, the voltage induced in the coil is 1.25 V.

You might be interested in

In the steady state 1.2 ✕ 1018 electrons per second enter bulb 1. There are 6.3 ✕ 1028 mobile electrons per cubic meter in tungs

bekas [8.4K]

Answer:

E=12.2V/m

Explanation:

To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

The equation is given by,

$V=\frac{I}{nAq}$

Where,

V= Drift Velocity

I= Flow of current

n= number of electrons

q = charge of electron

A = cross-section area.

For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is

$\frac{I}{q} = 1.2*10^{18}$

$A= 1.3*10^{-8}m^2$

$n=6.3*10^{28} e/m^3$

$\omicron{O} = 1.2*10^{-4}(m/s)(N/c)$ Mobility

We can find the drift velocity replacing,

$V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}$

$V= 1.465*10^-3m/s$

The electric field is given by,

$E= \frac{V}{\omicron{O}}$

$E=\frac{1.465*10^-3}{1.2*10^{-4}}$

$E=12.2V/m$

7 0

3 years ago

If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s then the KINETIC ENERGY of the CANNON will be ("E4" means

lilavasa [31]

Answer:

M V = m v conservation of momentum (Caps-cannon Small-projectile)

V = m / M * V = 2 / 2000 * 200 m/s = .2 m/s recoil velocity of cannon

KE = 1/2 M V^2 = 2000 / 2 kg * (.2 m/s)^2 = 40 kg m^2/s^2 = 40 J

6 0

3 years ago

Read 2 more answers

What information can be determined by observing dark lines within the spectrum of a given star?

Scrat [10]

I believe the answer is D.

6 0

2 years ago

Estimate the final temperature of a mole of gas at 200.0 atm and 19.0°C as it is forced through a porous plug to a final pressur

Yanka [14]

Answer : The final temperature of gas is 266.12 K

Explanation :

According to the Joule-Thomson experiment, it states that when a gas is expanded adiabatically from higher pressure region to lower pressure region, the change in temperature with respect to change in pressure at constant enthalpy is known as Joule-Thomson coefficient.

The formula will be:

$\mu_{J,T}=(\frac{dT}{dP})_H$

or,

$\mu_{J,T}=(\frac{dT}{dP})_H\approx \frac{\Delta T}{\Delta P}$

As per question the formula will be:

$\mu_{J,T}=\frac{T_2-T_1}{P_2-P_1}$ .........(1)

where,

$\mu_{J,T}$ = Joule-Thomson coefficient of the gas = $0.13K/atm$

$T_1$ = initial temperature = $19.0^oC=273+19.0=292.0K$

$T_2$ = final temperature = ?

$P_1$ = initial pressure = 200.0 atm

$P_2$ = final pressure = 0.95 atm

Now put all the given values in the above equation 1, we get:

$0.13K/atm=\frac{T_2-292.0K}{(0.95-200.0)atm}$

$T_2=266.12K$

Therefore, the final temperature of gas is 266.12 K

5 0

2 years ago

Why might the term red hot be misleading? (relating to stars)

Allushta [10]

red hot is hot, but other colours are even hotter. stars may be hotter than red hot.

there is also something calle the red shift.

3 0

2 years ago