Question

A circular loop of flexible iron wire has an initial circumference of 170 cm , but its circumference is decreasing at a constant rate of 12.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.800 T, which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.
(a) Find the emf induced in the loop at the instant when 9.0 s have passed.
(b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.

Answer 1

Answer:

after 9.00 s ; the magnitude of the induced emf = 0.000944 V

The induced current will be in the clockwise direction.

Explanation:

To find the circumference of the circle ; we use the expression:

C = 2πr

differentiating the above expression to determine the rate of change of the circumference; we have:

$\frac{dC}{dt} = 2 \pi \frac{dr}{dt}$

Here ; the rate of change of the circumference is $\frac{dC}{dt}$

Replacing 12.0 cm/s for $\frac{dC}{dt}$; we have :

$2 \pi \frac{dr}{dt} = - 12.0 cm/s ------ \ (equation \ 1 )$

However ;the area of the circle is expressed as:

A = πr²

Also; the rate of change of the area is determined as;

$\frac{dA}{dt} = d \frac{\pi r^2}{dt}$

$\frac{dA}{dt} = 2 \pi r \frac{dr}{dt} ------ \ (equation \ 2)$

Here; the rate of change of area is $\frac{dA}{dt}$

Replacing - 12.0 cm/s for $2 \pi r \frac{dr}{dt}$ in equation (2) ; we have:

$\frac{dA}{dt} =(-12.0 \ cm/s) r ------ \ (equation \ 3)$

Now; after 9.00 s ; the value of circumference of the loop is decreased by:

$C_f = ((12.0 \ cm/s )( \frac{1 \ m}{10^2 \ cm}))(9 \ s)$

= 1.08 m

The expression for the circumference of the loop after 9.00 s is:

$C = C_i - C_f$

Given that :

$C_i ( initial \ circumference) = 170 \ cm = 1.7 \ m$

C = ( 1.7 - 1.08) m

C = 0.62 m

Recall that :

C = 2πr

0.62 = 2×3.14 × r

r = $\frac{0.62}{2*3.14}$

r = 0.099 m

Replacing r = 0.099 m into equation (3)

$\frac{dA}{dt} =((-12.0 \ cm/s )( \frac{1 \ m}{10^2 \ cm}))(0.099 \ m)$

$\frac{dA}{dt} =-0.01188 m^2/s$

From Faraday's law, Induced emf (ε) is expressed as:

$\epsilon = - \frac{d \phi }{dt}$

and the magnetic flux is given as:

$\phi = BAcos \theta$

replacing the value of $\phi$ into above equation; we have:

$\epsilon = - \frac{d \ (BAcos \theta) }{dt}$

= $- B cos \theta \frac{d(A)}{dt}$

where θ = 0 ; B = 0.800 T and $\frac{dA}{dt}$ = -0.0118 m²/s

$\epsilon = -(0.800 \ T) (cos 0) (-0.0118 m^2/s)$

$\epsilon = 0.000944 \ V$  

Therefore; after 9.00 s ; the magnitude of the induced emf = 0.000944 V

b) The magnitude of id directing into the plane; However ; considering Lenz's law ; it states that the changes produced in the field will be opposed by the induced current. Thus ; it is found that the direction of the current will be in clockwise direction.

Answer 2

Answer:

(a) induced emf across the circular loop after 9 seconds is 9.484mV.

(b) When viewed from a point where the magnetic field is coming into the circular loop, the direction of the induced current in the loop is in clockwise direction.

Explanation:

(a)

The induced  emf around a circular is given by:

ϵ= - [dФ ÷ dt]

Where Ф is the magnetic influx passing through the area of the coil.

The induced emf in a coil is equal to the negative of the rate of change of magnetic flux times the number of turns in the coil.

When the coil experience a change in value with respect to time, then emf will be induced.

The surface area of the coop is constant indicating the induced emf force will be zero.

imputing the value of the magnetic flux in the formula given above, we have:

ϵ= -  [d BA ÷ dt] = - B [d A ÷ dt]  - A [d B ÷ dt]

Where magnetic flux, Ф =BA

For the circumference, which is given by: 2πr

and the circumference of the coil given above= 170 cm

Therefore the radius = 170 ÷ 2π

=27.07 cm

Since it is decreasing with respect to time at a rate of 12.0 cm/s, we differentiate the circumference with respect to time:

2π [dr ÷ dt] = - 12.0 cm/s

The negative sign shows there is a decrease in the circumference. Divide both side by 2π, then we have:

[dr ÷ dt] =1.909 cm/s

Integrating to get the value of the radius of the circular at any given time we have:

r=1.909 t + C

At t= 0, the initial radius of the coil is 27.07 cm.

Hence, we get 27.07= 0+C

Therefore C= 27.07

The Area = πr²

When we differentiate the area with respect to time, we have:

[d A ÷ dt] = 2πr [dr ÷ dt]

[d A ÷ dt] = 2πr x   -1.909

Imputing the radius, i.e r= -1.909 t + C, we have:

[d A ÷ dt] = 2πr (-1.909t +27.07)x   -1.909

Then we find how much the area would decrease in 9 seconds

[d A ÷ dt] = 2πr (-1.909x9 +27.07)x   -1.909 = 118.55 cm²/s

Converting this to meter per second by multiplying by 10∧ -4, we have

[d A ÷ dt] = 0.011855

Hence the emf after 9 seconds

=  -0.800 x  0.011855 =0.009484V = 9.484mV

Hence the induced emf across the circular loop after 9 seconds is 9.484mV.

(b)

The direction of the induced current is the same as the direction of the induced emf.

When viewed from a point where the magnetic field is coming into the circular loop, the direction of the induced current in the loop is in clockwise direction.