A single-turn circular loop of wire of radius 5.0 cm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.02500.0250-\text{s}s time interval, the magnitude of the field increases uniformly from 200 to 300 mT. Determine the magnitude of the emf induced in the loop


Answer 1

Given Information:  

time = Δt = 0.0250 seconds

Radius = r = 5 cm = 0.05 m

Change in Magnetic field = ΔB = (0.300 - 0.200) T

Number of turns = N = 1

Required Information:  

Magnitude of induced emf = ξ = ?  


Magnitude of induced emf = ξ = 3.141x10⁻² V


The EMF induced in a circular loop of wire in a changing magnetic field is given by  

ξ = -NΔΦ/Δt  

Where change in flux ΔΦ is given by


ΔΦ = ΔBπr²

ΔΦ = (0.300 - 0.200)*π*(0.05)²

ΔΦ = 7.854x10⁻⁴ T.m²

ξ = -NΔΦ/Δt  

ξ = -(1*7.854x10⁻⁴)/0.0250  

ξ = -3.141x10⁻² V

The negative sign is due to Lenz law.

Answer 2


-0.0314 V


Parameters given:

Initial magnetic field, Bini = 200 mT = 0.2T

Final magnetic field, Bfin = 300mT = 0.3 T

Number of turns, N = 1

Radius, r = 5 cm = 0.05 m

Time, t = 0.025 secs

Induced EMF is given as:

EMF = [-(Bfin - Bini) * N * pi * r²] / t

EMF = [-(0.3 - 0.2) * 1 * 3.142 * 0.05²] / 0.025

EMF = (-0.1 * 3.142 * 0.0025) / 0.025

EMF = -0.0314 V

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Convert this measurement
664.2 km=____cm


(664.2 km) · (1,000 m/km) · (100 cm/m) =

(664.2 · 1,000 · 100) (km·m·cm/km·m) =

66,420,000 cm

For metric conversion, you can remember this acronym for help:

King Henry died unusually drinking chocolate milk. Which stand for:

Kilo - unit * 1000

Hecto - unit * 100

Deca - unit * 10

Unit - unit * 1

Deci - unit * (1)/(10)

Centi - unit * (1)/(100)

Milli - unit * (1)/(100)

Kilometers and centimeters are five places apart apart, so you move the decimal point in 664.2 to the right five times, which means 664.2 km = 66420000 cm.

To avoid confusion on which direction to move the decimal point, imagine two shapes on each end of a scale. On each end, there is one large shape and one small shape. There has to be one of each on either side for it to balance. For this problem, a kilometer is a larger unit than a centimeter, so this means that the blank space needs to have a number greater than 664.2, or else the scale won't balance. Hope this helped.

Find the linear velocity of a point moving with uniform circular motion, if the point covers a distance s in the given amount of time t. s



The linear velocity is represented by the following expression: v = (s)/(t)


From Rotation Physics we know that linear velocity of a point moving with uniform circular motion is:

v = r\cdot \omega(Eq. 1)


r - Radius of rotation of the particle, measured in meters.

\omega - Angular velocity, measured in radians per second.

v - Linear velocity of the point, measured in meters per second.

But we know that angular velocity is also equal to:

\omega = (\theta)/(t)(Eq. 2)


\theta - Angular displacement, measured in radians.

t - Time, measured in seconds.

By applying (Eq. 2) in (Eq. 1) we get that:

v = (r\cdot \theta)/(t)(Eq. 3)

From Geometry we must remember that circular arc (s), measured in meters, is represented by:

s = r\cdot \theta

v = (s)/(t)

The linear velocity is represented by the following expression: v = (s)/(t)

If my weight on Earth is 140lbs, what is my mass?



63.57 kg


weight = 140 lbs

Let the mass  is m.

1 lbs = 4.45 N

The weight of an object is defined as the force with which our earth attracts the body towards its centre.

Weight is the product of mass of the body and the acceleration due to gravity of that planet.

W = m x g

On earth surface g = 9.8 m/s^2

Now convert lbs in newton

So, 140 lbs = 140 x 4.45 = 623 N

So, m x 9.8 = 623

m = 63.57 kg

Thus, the mass is 63.57 kg.

A 2.13-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N/m. The object is displaced 3.54 m to the right from its equilibrium position and then released, initiating simple harmonic motion. (a) What is the force (magnitude and direction) acting on the object 3.50 s after it is released



17.54N in -x direction.


Amplitude (A) = 3.54m

Force constant (k) = 5N/m

Mass (m) = 2.13kg

Angular frequency ω = √(k/m)

ω = √(5/2.13)

ω = 1.53 rad/s

The force acting on the object F(t) = ?

F(t) = -mAω²cos(ωt)

F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)

F(t) = -17.65 * cos (5.355)

F(t) = -17.57N

The force is 17.57 in -x direction

Light from a sodium lamp (λ = 589 nm) illuminates two narrow slits. The fringe spacing on a screen 150 cm behind the slits is 4.0 mm. What is the spacing (in mm) between the two slits?


The spacing between the two slits is 0.221mm.

The spacing  between the two slits is given as,

                    D=(\lambda L)/(y)

Where \lambda is wavelength, y is fringe spacing and L is length of screen.

Given that, \lambda=589nm,L=150cm,y=4mm

Substitute in above equation.

               D=(589*10^(-9)*150*10^(-2)  )/(4*10^(-3) )\n \nD=2.21*10^(-4) m\n\nD=0.221mm

Hence, the spacing between the two slits is 0.221mm.

Learn more about the sodium lamp here:


In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its speed is decreased linearly from 60 mph to 30 mph in 10 seconds. Calculate the theoretical maximum energy in kWh that can be recovered during this interval. Ignore all losses.



the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh


Given that;

weight of vehicle = 4000 lbs

we know that 1 kg = 2.20462


m = 4000 / 2.20462 =  1814.37 kg

Initial velocity V_(i) = 60 mph = 26.8224 m/s

Final velocity V_(f) = 30 mph = 13.4112 m/s

now we determine change in kinetic energy

Δk = (1)/(2)m(  V_(i)² - V_(f)² )

we substitute

Δk = (1)/(2)×1814.37( (26.8224)² - (13.4112)² )

Δk = (1)/(2) × 1814.37 × 539.5808

Δk = 489500 Joules

we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule


Δk = 489500 / 3.6 × 10⁶

Δk = 0.13597 ≈ 0.136 kWh

Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh