A coil of one loop is made from a wire of length L and there after a coil of two loops is made from same wire, then the ratio of magnetic field at the centre of coils will be
A. 1:4
B. 1:1
C. 1:8
D. 4:1
Answer
627k+ views
Hint: Place the values in the formula for length of wire by taking two different radii into consideration and on equating them we find the answer.
A magnetic field is a vector field that describes the magnetic influence on an electric charge of other moving charges or magnetised materials.
Complete step by step solution:
Write all the values which are provided,
Length of the wire is L
Now we can let the radius of first and second loop be $R_1$ and $R_2$ respectively,
Then,
According to the question,
\[L = 2\pi {R_1}\](for the first loop)
And,
\[L = 2 \times 2\pi {R_2}\](second loop)
Now equating we get,
\[2\pi {R_1} = 2 \times 2\pi {R_2}\]
Bringing \[{R_1}\]and \[{R_2}\]in one side we get,
\[\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{2\pi }}{{2 \times 2\pi }}\]
By further solving we get,
\[\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{1}{2}\]
Now we know that magnetic field at the centre of the coil of radius R is given by,
\[B = \dfrac{{{\mu _0}nI}}{{2R}}\]
Now for the first and second loop the magnetic field at the centre of the coil of radius \[{R_1}\]and \[{R_2}\]is
\[{B_1}\]=\[\dfrac{{{\mu _0}{n_1}I}}{{2{R_1}}}\]
And,
\[{B_2}\]=\[\dfrac{{{\mu _0}{n_2}I}}{{2{R_2}}}\]
Now,
\[\dfrac{{{B_1}}}{{{B_2}}} = \dfrac{{\dfrac{{{\mu _0}{n_1}I}}{{2{R_1}}}}}{{\dfrac{{{\mu _0}{n_2}I}}{{2{R_2}}}}}\]
Now cancelling out \[{\mu _0}\], 2 and I we get,
\[\dfrac{{{B_1}}}{{{B_2}}} = \dfrac{{{n_1}{R_2}}}{{{n_2}{R_1}}}\]
Now we know the values of \[{n_1} = \]1, \[{n_2} = \]2 and \[\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{1}{2}\]
Putting the values in the above equation we get,
\[\dfrac{{{B_1}}}{{{B_2}}} = \dfrac{1}{2} \times \dfrac{1}{2}\]
Which is equal to
\[\dfrac{{{B_1}}}{{{B_2}}} = \dfrac{1}{4}\]
It can be also written as,
\[{B_1}:{B_2} = 1:4\]
Therefore, the correct option is A.
Note: Students must remember that the radius of the two will not be same that is why we have two different radii i.e. R1 and R2.
A magnetic field is a vector field that describes the magnetic influence on an electric charge of other moving charges or magnetised materials.
Complete step by step solution:
Write all the values which are provided,
Length of the wire is L
Now we can let the radius of first and second loop be $R_1$ and $R_2$ respectively,
Then,
According to the question,
\[L = 2\pi {R_1}\](for the first loop)
And,
\[L = 2 \times 2\pi {R_2}\](second loop)
Now equating we get,
\[2\pi {R_1} = 2 \times 2\pi {R_2}\]
Bringing \[{R_1}\]and \[{R_2}\]in one side we get,
\[\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{2\pi }}{{2 \times 2\pi }}\]
By further solving we get,
\[\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{1}{2}\]
Now we know that magnetic field at the centre of the coil of radius R is given by,
\[B = \dfrac{{{\mu _0}nI}}{{2R}}\]
Now for the first and second loop the magnetic field at the centre of the coil of radius \[{R_1}\]and \[{R_2}\]is
\[{B_1}\]=\[\dfrac{{{\mu _0}{n_1}I}}{{2{R_1}}}\]
And,
\[{B_2}\]=\[\dfrac{{{\mu _0}{n_2}I}}{{2{R_2}}}\]
Now,
\[\dfrac{{{B_1}}}{{{B_2}}} = \dfrac{{\dfrac{{{\mu _0}{n_1}I}}{{2{R_1}}}}}{{\dfrac{{{\mu _0}{n_2}I}}{{2{R_2}}}}}\]
Now cancelling out \[{\mu _0}\], 2 and I we get,
\[\dfrac{{{B_1}}}{{{B_2}}} = \dfrac{{{n_1}{R_2}}}{{{n_2}{R_1}}}\]
Now we know the values of \[{n_1} = \]1, \[{n_2} = \]2 and \[\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{1}{2}\]
Putting the values in the above equation we get,
\[\dfrac{{{B_1}}}{{{B_2}}} = \dfrac{1}{2} \times \dfrac{1}{2}\]
Which is equal to
\[\dfrac{{{B_1}}}{{{B_2}}} = \dfrac{1}{4}\]
It can be also written as,
\[{B_1}:{B_2} = 1:4\]
Therefore, the correct option is A.
Note: Students must remember that the radius of the two will not be same that is why we have two different radii i.e. R1 and R2.
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