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# can you please give a full worked out solution for the question attached ?

Umakant biswal
5359 Points
4 years ago
@ tejaswini
apply the formula
e = i/ r
and e = blv sin theta
in which b is the magnetic field , v is the volume and l is the length of the structure …
and for direction u can use lenzes law .
i have solved your ques previously , kindly check it and tell in which step u have problem .
ALL THE BEST ..
Raman Mishra
67 Points
4 years ago
In the square configuration, area covered = l2
=> Initial flux = B.A = -Bl2 (considering area vector out of the page to be +ve)
For the area of the rhombus configuration, we have:
angle opposite to $\theta$ is also $\theta$ and the other two remaining angles are $\dpi{80} \pi -\theta$ each.
The diagonals of a rhombus are perpendicular to each other and  can be found using sine and cosine rule in the triangles formed by their intersection. Solving, we get:
$\dpi{80} d_{1}= 2l\sin \frac{\theta }{2}$  and     $\dpi{80} d_{2}=2l\cos \frac{\theta }{2}$
$\dpi{80} \therefore$  Area = $\dpi{80} \frac{1}{2}d_{1}d_{2} = l^{2}\sin \theta$
Hence, Final flux cut by the rhombus = -$\dpi{80} B l^{2}\sin \theta$
$\dpi{80} \therefore$ Change in flux =  -$\dpi{80} B l^{2}\sin \theta$ – (-Bl2) = $\dpi{80} Bl^{2}(1-\sin \theta )$
$\dpi{80} \therefore$ emf induced in the coil = e = $\dpi{80} -\frac{\Delta \Phi }{\Delta t} = -\frac{Bl(1-\sin \theta )}{\Delta t}$
Here, negative sign indicates that the induced emf opposes its cause according to the Lenz’s law.
Now, current induced in the coil = e/R  $\dpi{80} = -\frac{Bl(1-\sin \theta )}{R \Delta t }$
Since the area and hence flux(its direction out of the page) decrease current will be induced in the anticlockwise direction.
Ritika Singh
11 Points
3 years ago
Flux= B.A the emf is induced due to changing area.A=[ 1/2×(2lsintheta)×(lcostheta)]2A=l^2sin2thetaNow since the area is varying:dA/dt = (l^2cos2theta)dtheta/dt So d (flux)/dt=emf induced= B [dA/dt] = B [l^2cos2theta (dtheta/dt)]Now we have to calculate dtheta/dt:x= 2lsinthetadx/dt= 2lcostheta(dtheta/dt)2v=2lcostheta (dtheta/dt)v= lcostheta (dtheta/dt)dtheta/dt= v/lcosthetaTherefore on substituting this value in the equation of emf induced, you get the induced emf and hence apply:i= e/r