Question

ODBAC is a fixed rectangular conductor of negilible resistance (CO is not connnected) and OP is a conductor which rotates clockwise with an angular velocity ω (Figure). The entire system is in a uniform magnetic field B whose direction is along the normal to the surface of the rectangular conductor ABDC. The conductor OP is in electric contact with ABDC. The rotating conductor has a resistance of λ per unit length. Find the current in the rotating conductor, as it rotates by 180°.

Answer 1

When the conductor OP is rotated, then the rate of change of area and hence the rate of change of flux can be considered uniform from `0 < θ < pi/4; pi/4 < θ < (3pi)/4` and `(3pi)/4 < θ < pi/2`.

(i) Let us first assume the position of rotating conductor at time interval 

t = 0 to `t = pi/(4ω)` (or T/8)

The rod OP will make contact with the side BD. Let the length OQ of the contact after some time interval t such that `0 < t < pi/(4ω)` or `ω < t < T/8` be x.

The flux through the area ODQ is `phi_m = BA = B(1/2 xx QD xx OD) = B(1/2 xx l tan θ x l)`

⇒ `phi_m = 1/2 Bl^2 tan θ`, where θ = ωt

By applying Faraday's law of EMI,

Thus, the magnitude of the emf induced is `|ε| = |(dphi)/(dt)| = 1/2 Bl^2 ω sec^2 ωt`

The current induced in the circuit will be `I = ε/R` where, R is the resistance of the rod in contact.

Where, `R  oo  λ`

R = `λx = (λl)/(cos ωt)`

∴ `I = 1/2 (Bl^2ω)/(λl) sec^2 ωt cos ωt = (Blω)/(2λ cos ωt)`

(ii) Now let the rod OP will make contact with the side AB. And the length of OQ of the contact after some time interval t such that `pi/(4ω) < t < (3pi)/(4ω)` or `T//8 < t < (3T)/8` be x.

The flux through the area OQBD is `phi_m = (l^2 + 1/2 l^2/(2 tan theta))B`

Where, θ = ωt

 Thus, the magnitude of emf induced in the loop is 

`|ε| = |(dphi)/(dt)| = (Bl^2 ω sec^2 ωt)/(2 tan^2 ωt)`

The current induced in  the circuit is `I = ε/R = ε/(λx) = (ε sin ωt)/(λl) = 1/2 (Blω)/(λ sin ωt)`

(iii) When the flux through OQABD = `phi` = B.A

`phi = B.(2l^2 + 1/2 ly)`

`phi = B.(2l^2 + (l^2 tan ωt)/2)`  ......`[(tan (180 - theta) = y/l),((y = l(- tan theta)),(y = - l tan theta))]`

`(dphi)/(dt) = d/(dt) [2l^2 + l^2/2 (tan ωt)]B`

`- ε = 0 + (Bl^2)/2 + d/(dt) tan ωt`

`- ε = + (Bl^2ω)/2 sec^2 ωt = - (Bl^2ω)/(2 cos^2 ωt)`

`I = ε/R = ε/(λx)`

`I = (- Bl^2ω)/(2 cos^2  ωt) (cos  ωt)/(λ(-1))`  ......`[because l/x = cos (180 - θ) 1/x = - cos θ ⇒ x = (-1)/(cos ωt)]`

`I = (Blω)/(2λ cos ωt)`