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Question: A wire in the form of a circular loop of one turn carrying a current produces a magnetic field B at ...

A wire in the form of a circular loop of one turn carrying a current produces a magnetic field B at the center. If the same wire is looped into a coil of two turns and carrying the same current, the new value of magnetic field at the center is:
A. 3B
B. 5B
C. 4B
D. 2B

Explanation

Solution

The number of turns in the wire corresponds to the magnetic field. The more the number of turns, the more will be the magnetic field. We can consider turns as individual loops and directly find the magnetic field at the center of the loop by adding them individually. In short, for thin wires, the net magnetic field at the center is N times the field produced by the individual loop.

Formula used : dB=μi×dl4πr2dB=\dfrac { { \mu }_{ \circ }i\times dl }{ 4\pi r^{ 2 } } [ Statement of Biot-Savart law ]

Complete step-by-step answer :
For a single turn loop, the elementary magnetic field at a distance ‘r’ from a current ‘i’ carrying element ‘dl’ is given by Biot-Savart law:
dB=μi×dl4πr2dB=\dfrac { { \mu }_{ \circ }i\times dl }{ 4\pi r^{ 2 } }
For a circular loop, the distance from the centre of all the elements is ‘r’ hence distance is constant for all the elements.
For the net magnetic field, we’ll integrate the magnetic field.
dB=02πRμi×dl4πr2=μi4πr202πRdl=μi4πr2(2πR0)=μi2r\int { dB } =\int _{ 0 }^{ 2\pi R }{ \dfrac { { \mu }_{ \circ }i\times dl }{ 4\pi r^{ 2 } } } =\dfrac { { \mu }_{ \circ }i }{ 4\pi r^{ 2 } } \int _{ 0 }^{ 2\pi R }{ dl } =\dfrac { { \mu }_{ \circ }i }{ 4\pi r^{ 2 } } (2\pi R-0)=\dfrac { { \mu }_{ \circ }i }{ 2r } [ r=Rr=R]
Hence, Bnet=μi2rB_{net} = \dfrac { { \mu }_{ \circ }i }{ 2r }
Now, for N number of turns in a loop, B=N×μI2rB=N \times \dfrac { { \mu }_{ \circ }I }{ 2r }
Now, Given for 1 turn, the field is B:
B=μi2rB= \dfrac { { \mu }_{ \circ }i }{ 2r }
And for 2 turns and same current:
B2=2×μi2rB_{2} =2\times \dfrac { { \mu }_{ \circ }i }{ 2r }
But μi2r=B\dfrac { { \mu }_{ \circ }i }{ 2r } =B
So, B2=2×BB_{2} = 2 \times B
Hence B2=2BB_{2} = 2 B
D. option is correct.

Additional information : Biot savart’s law: In electromagnetism, Biot savart’s law is a law that relates a constant current flowing in a conducting wire with the magnetic field produced by it in its surrounding, at some distance. It relates the magnetic field to the magnitude, direction, length of the electric current.

Note :Students should practice these kinds of derivations as much as possible in order to remember them effectively. They also should visualize the direction of the magnetic field by applying the right hand thumb rule. In this question, r is constant. Hence we took it out of integration, otherwise, we have to make the distance as a function of the position of the element and integrate it likewise.