Question

A straight wire of length \[{\pi ^2}\] meter is carrying a current of 2A and the magnetic field due to it is measured at a point distance 1 cm from it. If the wire is to be bent into a circle and is to carry the same current as before, the ratio of the magnetic field at its centre to that obtained in the first case would be:
A. 50 : 1
B. 1 : 50
C. 100 : 1
D. 1 : 100

Answer 1

Hint: Since the wire is bent to form the circle of length l, the length of the wire will be equal to the circumference of the circle of radius R. Recall the expression for the magnetic field at a distance d from the straight current carrying wire and also express the magnetic field at the centre of the current carrying loop. Take the ratio of magnetic field at the centre of the circle and magnetic field at a distance 1 cm from the straight wire.

Formula used:
Magnetic field due to straight current carrying wire, \[B = \dfrac{{{\mu _0}i}}{{2\pi d}}\]
Here, \[{\mu _0}\] is the permeability of the free space, I is the current in the wire and d is the distance of the point P from the wire where the magnetic field is to be determined.
The magnetic field at the centre of the current carrying loop of the wire of radius R,
\[B' = \dfrac{{{\mu _0}i}}{{2R}}\]

Complete step by step answer:
We have the wire of length \[{\pi ^2}\] meter. This wire is bent to form the circle of length l. Therefore, the length of the wire will be equal to the circumference of the circle of radius R.
\[l = 2\pi R\]
\[ \Rightarrow R = \dfrac{l}{{2\pi }}\]
Substituting \[{\pi ^2}\] for l in the above equation, we get,
\[R = \dfrac{{{\pi ^2}}}{{2\pi }}\]
\[ \Rightarrow R = \dfrac{\pi }{2}\]
We have the expression for the magnetic field at a distance d from the straight current carrying wire,
\[B = \dfrac{{{\mu _0}i}}{{2\pi d}}\] …… (1)
Here, \[{\mu _0}\] is the permeability of the free space, I is the current in the wire and d is the distance of the point P from the wire where the magnetic field is to be determined.
We have the expression for the magnetic field at the centre of the current carrying loop of the wire of radius R,
\[B' = \dfrac{{{\mu _0}i}}{{2R}}\] …… (2)
Dividing equation (2) by equation (1), we get,
\[\dfrac{{B'}}{B} = \dfrac{{\dfrac{{{\mu _0}i}}{{2R}}}}{{\dfrac{{{\mu _0}i}}{{2\pi d}}}}\]
\[ \Rightarrow \dfrac{{B'}}{B} = \dfrac{{\pi d}}{R}\]
Substituting \[d = 1\,{\text{cm}} = 0.01\,{\text{m}}\] and \[R = \dfrac{\pi }{2}\] in the above equation, we get,
\[\dfrac{{B'}}{B} = \dfrac{{\pi \left( {0.01} \right)}}{{\dfrac{\pi }{2}}}\]
\[ \Rightarrow \dfrac{{B'}}{B} = 0.02\]
\[ \Rightarrow \dfrac{{B'}}{B} = \dfrac{1}{{50}}\]
\[ \therefore B':B = 1:50\]

So, the correct answer is option B.

Note: In the expression for the magnetic field at a certain point due to current carrying wire, d is not the radius of cross section the wire. One can solve the question by separately calculating the magnetic fields due to straight wire and circular loop. The radius of the circular loop is greater than the distance d since the radius is equal to\[\dfrac{\pi }{2}\].