Question

Figure shows a square frame of wire having a total resistance r placed coplanarly with a long, straight wire. The wire carries a current i given by i = i₀ sin ωt. Find (a) the flux of the magnetic field through the square frame, (b) the emf induced in the frame and (c) the heat developed in the frame in the time interval 0 to \[\frac{20\pi}{\omega}.\]

Answer 1

Let us consider an element of the loop of length dx at a distance x from the wire.

(a) Area of the element of loop A = adx

Magnetic field at a distance x from the wire, `(20 pi)/omega`

The magnetic flux of the element is given by

`d phi = (mu_0i xx adx)/(2 pi x)`

The total flux through the frame is given by

`phi = intd phi`

= `int_b^(a + b) (mu_0iadx)/(2 pi x)`

= `(mu_0ia)/(2 pi) ln [1 + a/b]`

(b) The e.m.f. induced in the frame is given by

`d/dt (mu_0ia)/(2 pi) ln [1 + a/b]`

= `(mu_0a)/(2 pi) ln [1 + a/b] d/dt (i_0 sin omegat)`

= `(mu_0ai_0omega cos omegat)/(2 pi) ln [1 + a/b]`

(c) The current through the frame is given by

`i = e/r`

= `(mu_0ai_0omega cos omega t)/(2 pi r) ln [1 + a/b]`

The heat developed in the frame in the given time interval can be calculated as:-

`H = i^2rt`

= `[(mu_0ai_0omega cos omega t)/(2 pi r) ln (1 + a/b)]^2 xx r xx t`

`H = (mu_0^2a^2i_0^2omega^2cos^2(omegat))/((2 pi r))ln^2(1 + a/b)t`

Using `t = (20 pi)/omega`, we get

Since cos²ωt averages to 1/2 over a full cycle:

`H = (mu_0^2a^2i_0^2omega^2)/(4pi^2r^2)ln^2(1 + a/b) * 1/2 * (20 pi)/omega`

`H = (mu_0^2 xx a^2 xx i_0 ^2 xx omega)/(4 pi xx r) ln^2(1 + a/b) * 10`

= `(5mu_0^2a^2i_0^2omega)/(pi r) ln^2 (1 + a/b)`