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Question: A coil of mean area \(500c{{m}^{2}}\) and having 1000 turns is held perpendicular to a uniform field...

A coil of mean area 500cm2500c{{m}^{2}} and having 1000 turns is held perpendicular to a uniform field of 0.4 gauss. The coil is turned through 1800{{180}^{0}} in 110\dfrac{1}{10} second. The average induced emf is
A.0.02V
B. 0.04V
C. 1.4V
D. 0.08V

Explanation

Solution

Hint : To find the answer to this question, first find the initial flux and the final flux after the rotation of the coil in the magnetic field. Emf can be given in terms of the change in flux per unit time. Obtain the change in flux and time interval and put it in the equation to find the induced emf.

Complete step by step solution :
Given, the area of the coil is, A=500cm2A=500c{{m}^{2}}
Converting it into SI unit i.e. in m2{{m}^{2}}
A=500cm2=500×10−4m2=5×10−2m2A=500c{{m}^{2}}=500\times {{10}^{-4}}{{m}^{2}}=5\times {{10}^{-2}}{{m}^{2}}
The uniform magnetic field is,
B=0.4G=0.4×10−4TB=0.4G=0.4\times {{10}^{-4}}T
Time taken to turn the coil by an angle of 1800{{180}^{0}} is,
t=110s=0.1st=\dfrac{1}{10}s=0.1s
Now, the magnetic flux through a coil placed in a uniform magnetic field is given by,
ϕ=NBAcos⁡θ\phi =NBA\cos \theta
Where, N is the number of turns in the coil, B is the magnetic field and A is the area of the coil.
θ\theta is the angle between the coil and the magnetic field.
Now, initially the magnetic flux through the coil is,
ϕ1=NBAcos⁡00 ϕ1=NBA \begin{aligned} & {{\phi }_{1}}=NBA\cos {{0}^{0}} \\\ & {{\phi }_{1}}=NBA \\\ \end{aligned}
Finally, the coils turn through an angle 1800{{180}^{0}}. So, the magnetic flux through the coil is given as,
ϕ2=NBAcos⁡1800 ϕ2=−NBA \begin{aligned} & {{\phi }_{2}}=NBA\cos {{180}^{0}} \\\ & {{\phi }_{2}}=-NBA \\\ \end{aligned}
Now, the change in magnetic flux us given as,
Δϕ=ϕ2−ϕ1 Δϕ=−NBA−NBA Δϕ=−2NBA \begin{aligned} & \Delta \phi ={{\phi }_{2}}-{{\phi }_{1}} \\\ & \Delta \phi =-NBA-NBA \\\ & \Delta \phi =-2NBA \\\ \end{aligned}
Taking the magnitude only, the change in magnetic flux through the coil is,
Δϕ=2NBA\Delta \phi =2NBA
Now, the induced emf in a coil moving in a magnetic field is given by the expression.
emf=ΔϕΔtemf=\dfrac{\Delta \phi }{\Delta t}
Putting the values on the above equation, we get,
emf=2NBAΔt emf=2×1000×0.4×10−4×5×10−20.1 emf=0.04V \begin{aligned} & emf=\dfrac{2NBA}{\Delta t} \\\ & emf=\dfrac{2\times 1000\times 0.4\times {{10}^{-4}}\times 5\times {{10}^{-2}}}{0.1} \\\ & emf=0.04V \\\ \end{aligned}
So, the induced emf on the coil is 0.04 V.

Note : The induced voltage in a circuit due to the change in magnetic flux is directly proportional to this rate of change of the magnetic flux through the circuit. It is known as the phenomenon of electromagnetic induction.