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
630.3k+ 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.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

What is cell theory Who formulated it class 11 biology CBSE

Phyllotaxy is the arrangement of ALeaflets BLeaves class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

The symbiotic association of fungi and algae is called class 11 biology CBSE

Cell theory was formulated by A Schleiden and Schwann class 11 biology CBSE

