
A small flat search coil of area with 20 \[c{{m}^{2}}\] closely wound turns is positioned normal to the field direction and then quickly snatched from the field region. The total charge flowing in the coil ( measured by a ballistic galvanometer connected to the coil )is 7.5 m. The resistance of the coil and galvanometer is 0.8 \[\Omega \]. The field strength of the magnet is:
A. 1.25 T
B. 0.50 T
C. 0.75 T
D. 2.10 T
Answer
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Hint: We have been asked to calculate the strength of magnet which means that we are being asked to calculate the magnetic field. We know that magnetic fields can be calculated by using Faraday's law which states the relation between magnetic flux, the magnetic field and the area of the loop. For that, we will need to calculate the magnetic flux. For that, we will be using Faraday’s law of induction.
Formula Used:
\[\varepsilon =-N\dfrac{d\phi }{dt}\]
Where,
N is number of turns of coil
\[\phi \] = magnetic flux
t is time
\[I=\dfrac{\varepsilon }{R}\] , Ohm’s law
\[{{\phi }_{i}}=B.A\]
B is the magnetic field
A is the area of the loop
Complete step by step answer:
We know from Ohm’s law that
\[I=\dfrac{\varepsilon }{R}\] ……………………… (1)
But, from Faraday’s law we know that
\[\varepsilon =-N\dfrac{d\phi }{dt}\] ……………………… (2)
Therefore, from (1) and (2) we can say that
\[I=\left( \dfrac{-N}{R} \right)\left( \dfrac{d\phi }{dt} \right)\]
On rearranging the terms
We get,
\[Idt=\left( \dfrac{-N}{R} \right)d\phi \] …………………… (3)
We know that
\[Q=\int{Idt}\] ……………….. (4)
Therefore, from (3) and (4)
We get,
\[Q=Idt=\left( \dfrac{-N}{R} \right)\int_{{{\phi }_{i}}}^{{{\phi }_{f}}}{d\phi }\]
After integrating
\[Q=\left( \dfrac{-N}{R} \right)\left[ {{\phi }_{f}}-{{\phi }_{i}} \right]\]
It is given that final flux is zero
Therefore,
\[Q=\left( \dfrac{N}{R} \right){{\phi }_{i}}\] ………………. (5)
Now, we know that
\[{{\phi }_{i}}=B.A\] …………………… (6)
From (5) and (6)
We get,
\[Q=\dfrac{N\times B\times A}{R}\]
On rearranging the above equation
We get,
\[B=\dfrac{Q\times R}{N\times A}\]
Therefore, after substituting the given values
We get,
\[B=\dfrac{7.5\times {{10}^{-3}}\times 0.8}{20\times 4\times {{10}^{-4}}}\]
Therefore,
\[B=0.75T\]
Therefore, the correct answer is option C.
Note:
Faraday’s law of induction describes how an electric current generates a magnetic field and the reverse how a change in magnetic field will produce electric current. The component of magnetic field through a surface is known as magnetic flux through the same surface.
Formula Used:
\[\varepsilon =-N\dfrac{d\phi }{dt}\]
Where,
N is number of turns of coil
\[\phi \] = magnetic flux
t is time
\[I=\dfrac{\varepsilon }{R}\] , Ohm’s law
\[{{\phi }_{i}}=B.A\]
B is the magnetic field
A is the area of the loop
Complete step by step answer:
We know from Ohm’s law that
\[I=\dfrac{\varepsilon }{R}\] ……………………… (1)
But, from Faraday’s law we know that
\[\varepsilon =-N\dfrac{d\phi }{dt}\] ……………………… (2)
Therefore, from (1) and (2) we can say that
\[I=\left( \dfrac{-N}{R} \right)\left( \dfrac{d\phi }{dt} \right)\]
On rearranging the terms
We get,
\[Idt=\left( \dfrac{-N}{R} \right)d\phi \] …………………… (3)
We know that
\[Q=\int{Idt}\] ……………….. (4)
Therefore, from (3) and (4)
We get,
\[Q=Idt=\left( \dfrac{-N}{R} \right)\int_{{{\phi }_{i}}}^{{{\phi }_{f}}}{d\phi }\]
After integrating
\[Q=\left( \dfrac{-N}{R} \right)\left[ {{\phi }_{f}}-{{\phi }_{i}} \right]\]
It is given that final flux is zero
Therefore,
\[Q=\left( \dfrac{N}{R} \right){{\phi }_{i}}\] ………………. (5)
Now, we know that
\[{{\phi }_{i}}=B.A\] …………………… (6)
From (5) and (6)
We get,
\[Q=\dfrac{N\times B\times A}{R}\]
On rearranging the above equation
We get,
\[B=\dfrac{Q\times R}{N\times A}\]
Therefore, after substituting the given values
We get,
\[B=\dfrac{7.5\times {{10}^{-3}}\times 0.8}{20\times 4\times {{10}^{-4}}}\]
Therefore,
\[B=0.75T\]
Therefore, the correct answer is option C.
Note:
Faraday’s law of induction describes how an electric current generates a magnetic field and the reverse how a change in magnetic field will produce electric current. The component of magnetic field through a surface is known as magnetic flux through the same surface.
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