0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

# At time $$t = 0$$ magnetic field of $$1000$$ Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to $$500$$ Gauss, in nest $$5s$$, then induced EMF in the loop is:

A
$$48\mu V$$
B
$$36\mu V$$
C
$$28\mu V$$
D
$$56\mu V$$
Solution
Verified by Toppr

#### Correct option is C. $$56\mu V$$Area of loop $$AEBCFDA =$$ area of rectangle $$ABCD \ -$$ Area of triangle $$ABE\ -$$ Area of triangle $$DCF$$ $$\therefore$$ Required area $$=16cm \times 4 cm - \dfrac{1}{2}\times 4cm \times 2cm - \dfrac{1}{2}\times 4cm \times 2cm$$$$= 64cm^2 - 4cm^2 - 4cm^2 = 56cm^2$$Magnetic flux passing through an area is given by $$\displaystyle \phi = \int B.dA = \int BdA \cos \theta$$. Here, $$\theta = 0$$ and $$B$$ is constant throughout the area.$$\therefore \phi_{initial} = B_{initial}.A = 1000 \, Gauss \times 56cm^2$$$$= 56000 \, gauss\, cm^2$$$$= 5.6\times 10^{-4}Wb$$$$\begin{bmatrix} 1 \, Gauss = 10^{-4}T \, and \, 1cm^2 = 10^{-4}m^2\\ \therefore 1\, Gauss\, cm^2 = 10^{-8}Tm^2 = 10^{-8}Wb \end{bmatrix}$$Also, $$\phi _{final} = B_{final}. A = 500\, Gauss \times 56cm^2$$$$= 28000\, Gauss\, cm^2 = 2.8\times 10^{-4}Wb$$Since the flux changes linearly,$$\in = \left|\dfrac{d\phi}{dt}\right| = \dfrac{|\phi_{final} - \phi_{initial}|}{\Delta t} = \dfrac{(5.6-2.8)\times 10^{-4}}{5}V$$$$= 56\times 10^{-6}V = 56\mu v$$

14
Similar Questions
Q1

At time $t = 0$ magnetic field of $1000 Gauss$ is passing perpendicularly through the area defined by the closed-loop shown in the figure. If the magnetic field reduces linearly to $500 Gauss$ , in the next $5 s$, then induced $EMF$ in the loop is:

16am
View Solution
Q2
At time t=0 magnetic field of 1000 G is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 G, in the next 5 s, then induced EMF in the loop is:

View Solution
Q3
The flux of magnetic field through a closed conducting loop of resistance 0.4 changes with time according to the equation ϕ = 0.20t2 + 0.40t + 0.60 where t is time in seconds. Find (i) the induced emf at t = 2s. (ii) the average induced emf in t = 0 to t = 5 s. (iii) change passed through the loop in t = 0 to t = 5s. (iv) average current in time interval t = 0 to t = 5s (v) heat produced in t = 0 to t = 5s.
View Solution
Q4
96.The flux of magnetic field through a closed conducting loop of resistance 0.4 changes with time according to the equation = 0.20t2 + 0.40t + 0.60 where t is time in seconds. Find (i) the induced emf at t = 2s. (ii) the average induced emf in t = 0 to t = 5 s. (iii) charge passed through the loop in t = 0 to t = 5s (iv) average current in time interval t = 0 to t = 5 s (v) heat produced in t = 0 to t = 5s.
View Solution
Q5
In a uniform magnetic field. If a rod moves then Emf is induced. But if a closed loop moves in a plane perpendicular to uniform magnetic field then Emf induced will be 0?
View Solution