El6M.8 Two concentric circular coils lie on a tabletop. The outer coil has a rad

Question

El6M.8 Two concentric circular coils lie on a tabletop. The outer coil has a radius ro = 25 cm, consists of N = 100 turns of wire, and conducts a sinusoidal 60-Hz alternating current with a maximum value of io-5.0 A. The inner coil is much smaller, with a radius of r- 1.0 cm. It has a single turn (and so is just a loop) with a resistance of R 0.01 (a) What approximations do we have to make to calculate the emf and current induced in the small loop? (Hint: Any current flowing in the small loop will create a magnetic field. Why does this complicate things?) (b) Argue that the induced current in the small loop is 90 out of phase with the current in the large coil. (c) If the current in the large coil is going through zero on the way to becoming clockwise, in what direction does the current in the small loop flow? (d) Calculate the maximum emf and the maximum current flowing in the small loop (e) Argue on the basis of this answer that one of the approximations you made in part (a) is valid

Explanation / Answer

Radius of the outer coil, ro = 25 cm

Number of turns in outer coil, N = 100

Current in outer Coil, i(t) = io Sin(2*pi*f*t)

Where, io = 5.0 A, and f = 60 Hz.

radius of inner coil, r = 1 cm,

it has a single turn so it is just a loop.

Its resistance, R = 0.01 Ohm

Part A:

The time varying current will produce time varying magnetic field at the center of the coil. Hence flux of the magnetic field through the inner coil will change.

From Faraday's law, we know that Changing magnetic flux produces induced emf and current in the closed loop.

Now, to clculate the induced emf and current in the small loop, let us make the assumptions as below...

[i]

ro >> r, so that the entire inner loop can be considered at the center of the outer coil and magnetic field throughout the inner loop

B(t) = uo*i(t)/2r ------------->[1] {here, u = mu, magnetic permeability}

[ii]

neglecting any secondary magnetic field due to the induced current in the inner loop, as the secondary magnetic field is much smaller than the magnetic field due to current in outer loop.

[iii]

Both the coils are Co-planar i.e., 100 turns have negligible width.

Part B:

From Faraday's law, induced emf, e = - dQ/dt [here, Q = phi = magnetic flux]

flux, Q = B.A = {uo*N*i(t)/[2ro]}*pi*r2 through inner loop.

or

Q = {uo*N*io*pi*r2/[2ro]}*Sin(2*pi*f*t) -------> due to N turns

magnitude of induced emf, |e| = |dQ/dt| = {uo*N*io*pi*r2/[2ro]}*2*pi*f*Cos(2*pi*f*t)

induced current, id = |e|/R

id = |e|/R = {uo*N*io*pi2*r2/[Rro]}*f*Cos(2*pi*f*t)

OR

id = |e|/R = {uo*N*io*pi2*r2/[Rro]}*f*Sin(2*pi*f*t +90o) ------------>[3]

But current in large coil, i = io Sin(2*pi*f*t) ------------------------>[4]

From eqn [3]&[4], we conclude that induced current is 90o out of phase with current in large coil.

Hence Proved.

Part C:

From Lenz's law, we know that the induced current appears in such a way that it can oppose the cause, i.e., the causing magnetic field.

When current in large coil is in clockwise direction, the magnetic field is downward, from right hand rule. Then the induced current in small loop will be counterclock wise so that magnetic field due to it can oppose the causing magnetic field.

Part D:

From eqn[3], id = id,o Cos(2*pi*f*t)

where, id,o = {uo*N*io*pi2*r2/[Rro]}

Substituting the values given, we get  id,o = 4800*pi3*10-7 A

Now, id(max) = +  4800*pi3*10-7 A for Cos(2*pi*f*t) = 1 &

id(min) = - 4800*pi3*10-7 A for Cos(2*pi*f*t) = -1

and

max emf =  +  4800*pi3*10-9 V

min emf =  - 4800*pi3*10-9 V

Part E:

Now,

Secondary magnetic field due to id in small loop Bs = uoid/2r = 4*pi*10-7*4800*pi3*10-7 A /2*10-2 = 9.35*10-11 T

which is very small in accordance to the assumption made in [ii] from Part A.

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.