“The object of a New Year is not that we should have a new year. It is that we should have a new soul.”

– G. K. Chesterton

*Happy New Year…*

A free response practice question involving electromagnetism was given to you in the post dated 30

^{th}December 2009. As promised, I give below a model answer along with the question:

A rectangular wire loop of length *ℓ* and breadth *b *having* *negligible resistance is arranged in the plane of an infinitely long straight vertical wire as shown, with the longer sides parallel to the wire. The wire loop contains a capacitor of capacitance C. A steady current ‘*I*’ flows upwards through the straight wire. Now answer the following questions:

(a) Calculate the magnetic flux through the wire loop.

(b) The loop is rotated through 180º about a central axis (fig.) which is parallel to the straight wire. If the time taken for this rotation is ∆*t*, determine the magnitude of the emf induced in the loop.

(c) The loop is kept stationary and instead of the steady current *I*, a current ‘*i*’* *varying with time *t *as *i* = *I*_{m }sin *ωt* (where *I*_{m} and *ω* are constants) is passed through the straight wire. Calculate the maximum value of the emf induced in the wire loop.

(d) Determine the maximum current induced in the wire loop under the conditions mentioned in part (c) above.

**(a)** The magnetic field *B *(produced by the straight current carrying wire) at the loop varies with distance as

*B = *μ_{0}*I/*2π*r*** **where ‘

*r*’ is the distance of the point from the straight wire.

The magnetic field at the wire loop is directed *normally* in to the plane of the loop. The magnetic flux through the entire wire loop can be found considering strips parallel to the straight wire. One such strip of width *dr* at distance *r* is shown in the figure. This strip has area *dA = ℓdr* and hence the magnetic flux through this strip is given by

*dФ =BdA = *(μ_{0}*I/*2π*r*)*ℓdr*

The magnetic flux *Ф *through the entire wire loop is obtained by integrating the above expression between limits *r = b* and *r = *2*b*.

Therefore, *Ф =* (μ_{0}*Iℓ/*2π) * _{b}*∫

^{2b}(

*dr*/

*r*) = (μ

_{0}

*Iℓ/*2π)[ln(2

*b*) – ln(

*b*)]

Or, *Ф =* (μ_{0}*Iℓ/*2π) ln(2) = μ_{0}*Iℓ* ln(2)*/*2π

**(b)** When the wire loop is rotated through 180º, the magnetic flux through the loop changes from *Ф* to –*Ф* so that the change of flux is 2*Ф*. The magnitude of the emf *Ф/*∆*t* = 2×μ_{0}*Iℓ* ln(2)*/*2π∆*t* = μ_{0}*Iℓ* ln(2)*/*π∆*t*

**(c)** The magnetic flux linked with the wire loop when a current *i* = *I*_{m }sin *ωt* flows through the straight wire is obtained by replacing *I *by (*I*_{m }sin *ωt*) in the expression for *Ф* obtained in part (b) above.

Thus *Ф =* μ_{0}* *(*I*_{m }sin *ωt*)*ℓ* ln(2)*/*2π

The magnitude of the emf *dФ/dt*.

[The * dФ/dt*, the negative sign appearing because of Lenz’s law. We ignore the negative sign since we are interested in the magnitude of the emf].

Therefore, *V *= [μ_{0}*I*_{m}*ω**ℓ* ln(2)_{ }cos *ωt*]*/*2π

The maximum value of _{0}*I*_{m}*ω**ℓ* ln(2)_{ }*/*2π, appropriate to the maximum value of 1 for cos* ωt*.

**(d) **The charge *Q *on the capacitor because of the

*Q = CV = *[*C*μ_{0}*I*_{m}*ω*ℓ ln(2)_{ }cos *ωt*]*/*2π

The *i*_{ind}* *in the wire loop is given by

*i*_{ind} = *dQ/dt = *[*C*μ_{0}*I*_{m}*ω*^{2}*ℓ* ln(2)(–_{ }sin *ωt*)]*/*2π

The maximum value of induced current is *C*μ_{0}*I*_{m}*ω*^{2}*ℓ* ln(2)_{ }*/*2π.

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