
The magnetic flux of $500\mu Wb$ passing through a 200 turn coil is reversed in$20\times {{10}^{-3}}$ seconds. The average e.m.f. induced in the coil in volt is
A. 2.5
B. 5.0
C. 7.5
D. 10.0
Answer
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Hint: We are given the magnitude of magnetic flux in the question and we know that emf is only induced when there is a change in magnetic flux. Thus we could first find the change in flux due to its reversal. Then you could recall the expression for induced emf in terms of change in flux and substitute accordingly to get the answer.
Formula used:
Expression for emf,
$\varepsilon =-n\dfrac{\Delta \phi }{\Delta t}$
Complete answer:
In the question, we are given a magnetic flux of$500\mu Wb$ that is passing through a coil of 200 turns. This flux is reversed in the time interval of $20\times {{10}^{-3}}s$ and we asked to find the average e.m.f. induced in the coil in volts. So,
$n=200$ ……………………………………… (1)
${{\phi }_{1}}=500\mu Wb$
As the flux is just reversed, the direction of the flux is only changed with the same magnitude,
${{\phi }_{2}}=-500\mu Wb$
So, change in flux is given by,
$\Delta \phi ={{\phi }_{2}}-{{\phi }_{1}}$
$\Delta \phi =\left( -500-500 \right)\mu Wb$
$\therefore \Delta \phi =-1000\times {{10}^{-6}}Wb$ ……………………………………….. (2)
Time interval is given in the question as,
$\Delta t=20\times {{10}^{-3}}s$ …………………………………………….. (3)
Now, let us recall the expression for average emf induced in a coil due to the change in flux,
$\varepsilon =-n\dfrac{\Delta \phi }{\Delta t}$ …………………………………………….. (4)
Now, we could directly substitute (1), (2) and (3) in (4), we get,
$\varepsilon =-200\times \dfrac{-1000\times {{10}^{-6}}}{20\times {{10}^{-3}}}$
$\therefore \varepsilon =10V$
Therefore, we found the average e.m.f. induced in the coil in volt to be 10V.
Hence, option D is found to be the right answer.
Note:
You may have noticed the negative sign in the expression of induced emf. This negative sign represents the direction of current in a closed loop. Also, from the expression we see that the induced emf can be increased by increasing the number of turns in the coil as the induced emf is seen to be directly proportional to the number of turns.
Formula used:
Expression for emf,
$\varepsilon =-n\dfrac{\Delta \phi }{\Delta t}$
Complete answer:
In the question, we are given a magnetic flux of$500\mu Wb$ that is passing through a coil of 200 turns. This flux is reversed in the time interval of $20\times {{10}^{-3}}s$ and we asked to find the average e.m.f. induced in the coil in volts. So,
$n=200$ ……………………………………… (1)
${{\phi }_{1}}=500\mu Wb$
As the flux is just reversed, the direction of the flux is only changed with the same magnitude,
${{\phi }_{2}}=-500\mu Wb$
So, change in flux is given by,
$\Delta \phi ={{\phi }_{2}}-{{\phi }_{1}}$
$\Delta \phi =\left( -500-500 \right)\mu Wb$
$\therefore \Delta \phi =-1000\times {{10}^{-6}}Wb$ ……………………………………….. (2)
Time interval is given in the question as,
$\Delta t=20\times {{10}^{-3}}s$ …………………………………………….. (3)
Now, let us recall the expression for average emf induced in a coil due to the change in flux,
$\varepsilon =-n\dfrac{\Delta \phi }{\Delta t}$ …………………………………………….. (4)
Now, we could directly substitute (1), (2) and (3) in (4), we get,
$\varepsilon =-200\times \dfrac{-1000\times {{10}^{-6}}}{20\times {{10}^{-3}}}$
$\therefore \varepsilon =10V$
Therefore, we found the average e.m.f. induced in the coil in volt to be 10V.
Hence, option D is found to be the right answer.
Note:
You may have noticed the negative sign in the expression of induced emf. This negative sign represents the direction of current in a closed loop. Also, from the expression we see that the induced emf can be increased by increasing the number of turns in the coil as the induced emf is seen to be directly proportional to the number of turns.
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