Question

A circular loop of flexible iron wire has an initial circumference of 165cm , 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 oriented perpendicular to the plane of the loop of magnitude 0.500 T Part A Find the emf induced in the loop, at the instant when 9.0s have passed. And find the direction

Answer 1

Answer:

emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise

Explanation:

Given data

initial circumference = 165 cm

rate = 12.0 cm/s

magnitude = 0.500 T

tome = 9 sec

to find out

emf induced and direction

solution

we know emf in loop is - d∅/dt    ........1

here ∅ = ( BAcosθ)

so we say angle is zero degree and magnetic filed is uniform here so that

emf = - d ( BAcos0) /dt

emf = - B dA /dt     ..............2

so  area will be

dA/dt = d(πr²) / dt

dA/dt = 2πr dr/dt

we know 2πr = c,

r = c/2π = 165 / 2π

r  = 26.27 cm

c is circumference so from equation 2

emf = - B 2πr dr/dt    ................3

and

here we find rate of change of radius that is

dr/dt = 12/2π = 1.91  $10^{-2}$cm/s

so when 9.0s have passed that radius of coil = 26.27 - 191 (9)

radius = 9.08 $10^{-2}$ cm

so now from equation 3 we find emf

emf = - (0.500 )  2π(9.08 $10^{-2}$ )   1.91  $10^{-2}$

emf = - 0.005445

and magnitude of emf = 0.005445 V

so

emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise

Answer 2

The emf induced in the loop, at the instant when 9.0s has passed and the direction are;

EMF = 0.00544 V with the direction being clockwise.

We are given;

Initial circumference; C = 165 cm = 1.65 m

Rate of change of circumference; dC/dt = 12 cm/s = 0.12 m/s

Magnetic field; B = 0.5 T

From Faraday’s law of electromagnetic induction, the EMF is given as;

emf = d(BA)/dt

Where A is area of coil.

Differentiating gives;

EMF = B × dA/dt

Formula for area is;

A = πr²

differentiating with respect to t gives;

dA/dt = 2πr(dr/dt)

Putting 2πr(dr/dt) for dA/dt into the emf equation gives;

EMF = B × 2πr(dr/dt)

Now, formula for circumference is C = 2πr

Thus;

Thus; r = c/2π = 1.65/2π

r = 0.2626 m

After 9 s, C = dC/dt × 9

C = 0.12 × 9

C = 1.08 m

Thus; r = 1.08/2π

r' = 0.1719 m

After time of t = 9s

dr/dt = (0.1719 - 0.2626)/9

dr/dt = −0.010078 cm/s

Thus;

EMF = B × 2πr(dr/dt)

Plugging in the relevant values;

EMF = 0.5 × (2π × 0.1719) × (-0.010078)

EMF = −0.00544 V

We are dealing with magnitude and so we will take the absolute value which is;

EMF = 0.00544 V

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