Answer
397.8k+ views
Hint:In the above question, we have to find the total magnetic field. Then firstly, finding the magnetic field along the straight wire and then extending the circular loop and then again, the magnetic field of the straight wire. After adding all the magnetic fields, we got the final answer.
Complete step by step answer:
The total magnetic field will be given by adding the magnetic field of straight line and the magnetic field of the circular loop and again the magnetic field of straight line.Magnetic field can be calculated by the Biot savart’s law whose expression be given by,
$dB = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{Idr \times r}}{{{r^3}}}$
Magnetic field of straight wire be given by the formula,
$B = \dfrac{{{\mu _0}I}}{{4\pi r}}\left( {sin{\theta _1} + \sin {\theta _2}} \right)$
where ${\theta _{1,}}{\theta _2}$ are the angles corresponding to the two ends of the wire.
Now, the magnetic field along AB,
$
{B_1} = \dfrac{{{\mu _0}I}}{{4\pi R}}\left( {sin\dfrac{\pi }{2} - \sin \dfrac{\pi }{4}} \right) \otimes \\
\Rightarrow{B_1} = \dfrac{{{\mu _0}I}}{{4\pi R}}\left( {1 - \dfrac{1}{{\sqrt 2 }}} \right) \otimes \\
$ $ \ldots \ldots \left( 1 \right)$
Now, the magnetic field along the circular loop
${B_2} = \dfrac{{{\mu _0}I}}{{2R}} \odot $ $ \ldots \ldots \left( 2 \right)$
Magnetic field along the wire BC,
$
{B_3} = \dfrac{{\mu {}_0I}}{{4\pi R}}\left( {\sin \dfrac{\pi }{2} + \sin \dfrac{\pi }{4}} \right) \odot \\
\Rightarrow{B_3} = \dfrac{{{\mu _0}I}}{{4\pi R}}\left( {1 + \dfrac{1}{{\sqrt 2 }}} \right) \odot \\
$ $ \ldots \ldots \left( 3 \right)$
Now adding all the above marked equations:
$\overrightarrow B = {B_1} + {B_2} + {B_3}$
$
\Rightarrow\overrightarrow B = \left( {\dfrac{{{\mu _0}I}}{{2R}} + \dfrac{{2{\mu _0}I}}{{4\pi R}}\dfrac{1}{{\sqrt 2 }}} \right) \odot \\
\therefore\overrightarrow B = \dfrac{{{\mu _0}I}}{{2R}}\left( {1 + \dfrac{1}{{\pi \sqrt 2 }}} \right) \odot \\ $
Hence, the correct option is B.
Note:The value of proportionality is $\dfrac{{{\mu _0}}}{{4\pi }} = {10^{ - 7}}N{A^{ - 2}}$.Biot savart law is a vector quantity which contains both magnitude and direction.$ \otimes $ is the inward direction of the field and $ \odot $ is the outward direction of the field.
Complete step by step answer:
The total magnetic field will be given by adding the magnetic field of straight line and the magnetic field of the circular loop and again the magnetic field of straight line.Magnetic field can be calculated by the Biot savart’s law whose expression be given by,
$dB = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{Idr \times r}}{{{r^3}}}$
Magnetic field of straight wire be given by the formula,
$B = \dfrac{{{\mu _0}I}}{{4\pi r}}\left( {sin{\theta _1} + \sin {\theta _2}} \right)$
where ${\theta _{1,}}{\theta _2}$ are the angles corresponding to the two ends of the wire.
Now, the magnetic field along AB,
$
{B_1} = \dfrac{{{\mu _0}I}}{{4\pi R}}\left( {sin\dfrac{\pi }{2} - \sin \dfrac{\pi }{4}} \right) \otimes \\
\Rightarrow{B_1} = \dfrac{{{\mu _0}I}}{{4\pi R}}\left( {1 - \dfrac{1}{{\sqrt 2 }}} \right) \otimes \\
$ $ \ldots \ldots \left( 1 \right)$
Now, the magnetic field along the circular loop
${B_2} = \dfrac{{{\mu _0}I}}{{2R}} \odot $ $ \ldots \ldots \left( 2 \right)$
Magnetic field along the wire BC,
$
{B_3} = \dfrac{{\mu {}_0I}}{{4\pi R}}\left( {\sin \dfrac{\pi }{2} + \sin \dfrac{\pi }{4}} \right) \odot \\
\Rightarrow{B_3} = \dfrac{{{\mu _0}I}}{{4\pi R}}\left( {1 + \dfrac{1}{{\sqrt 2 }}} \right) \odot \\
$ $ \ldots \ldots \left( 3 \right)$
Now adding all the above marked equations:
$\overrightarrow B = {B_1} + {B_2} + {B_3}$
$
\Rightarrow\overrightarrow B = \left( {\dfrac{{{\mu _0}I}}{{2R}} + \dfrac{{2{\mu _0}I}}{{4\pi R}}\dfrac{1}{{\sqrt 2 }}} \right) \odot \\
\therefore\overrightarrow B = \dfrac{{{\mu _0}I}}{{2R}}\left( {1 + \dfrac{1}{{\pi \sqrt 2 }}} \right) \odot \\ $
Hence, the correct option is B.
Note:The value of proportionality is $\dfrac{{{\mu _0}}}{{4\pi }} = {10^{ - 7}}N{A^{ - 2}}$.Biot savart law is a vector quantity which contains both magnitude and direction.$ \otimes $ is the inward direction of the field and $ \odot $ is the outward direction of the field.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)