
Magnetic field due to current carrying conductor depends on:
A. Current flowing through the conductor
B. Distance from the conductor
C. Length of the conductor
D. (A) and (B)
Answer
520.8k+ views
Hint: According to Biot-Savart’s law, the magnetic field at any point due to a current element \[Id\overset{\to }{\mathop{l}}\,\] is given by:
\[d\overset{\to }{\mathop{B}}\,=\dfrac{{{\mu }_{o}}}{4\pi }\dfrac{Id\overset{\to }{\mathop{l}}\,\times \overset{\to }{\mathop{R}}\,}{{{R}^{3}}}\]
Where \[\overset{\to }{\mathop{R}}\,\] is the vector from the current element to the point of observation and the constant \[{{\mu }_{o}}\]is called the permeability of free space.
Complete step by step solution:
From Biot-Savart’s law, the magnetic field at any point due to a current carrying wire is
\[\overset{\to }{\mathop{B}}\,=\dfrac{{{\mu }_{o}}}{4\pi }\int{\dfrac{Id\overset{\to }{\mathop{l}}\,\times \overset{\to }{\mathop{R}}\,}{{{R}^{3}}}}\]
Where the integration is over the whole current path.
Thus, magnitude of the magnetic field at any point is proportional to the current I, the element length dl and inversely proportional to the square of the distance R of the point from the current element.
Hence, the magnetic field due to the current carrying conductor depends on the current flowing through the conductor and the distance from the conductor.
Therefore, option D is the correct answer.
Additional information:
In 1820, Hans Oersted, a Danish physics teacher, hypothesized that the current-carrying conductor produces a magnetic field around it and the direction of magnetic field depends on the direction of current.
Note: The direction of magnetic field will be the direction of the cross product \[d\overset{\to }{\mathop{l}}\,\times \overset{\to }{\mathop{R}}\,\]. It is represented by the right hand rule. The magnetic field inside the conductor is zero, while on the surface, the magnetic field is perpendicular to both the current density and the surface normal.
The Biot-Savart law is applicable only for steady currents.
\[d\overset{\to }{\mathop{B}}\,=\dfrac{{{\mu }_{o}}}{4\pi }\dfrac{Id\overset{\to }{\mathop{l}}\,\times \overset{\to }{\mathop{R}}\,}{{{R}^{3}}}\]
Where \[\overset{\to }{\mathop{R}}\,\] is the vector from the current element to the point of observation and the constant \[{{\mu }_{o}}\]is called the permeability of free space.
Complete step by step solution:
From Biot-Savart’s law, the magnetic field at any point due to a current carrying wire is
\[\overset{\to }{\mathop{B}}\,=\dfrac{{{\mu }_{o}}}{4\pi }\int{\dfrac{Id\overset{\to }{\mathop{l}}\,\times \overset{\to }{\mathop{R}}\,}{{{R}^{3}}}}\]
Where the integration is over the whole current path.
Thus, magnitude of the magnetic field at any point is proportional to the current I, the element length dl and inversely proportional to the square of the distance R of the point from the current element.
Hence, the magnetic field due to the current carrying conductor depends on the current flowing through the conductor and the distance from the conductor.
Therefore, option D is the correct answer.
Additional information:
In 1820, Hans Oersted, a Danish physics teacher, hypothesized that the current-carrying conductor produces a magnetic field around it and the direction of magnetic field depends on the direction of current.
Note: The direction of magnetic field will be the direction of the cross product \[d\overset{\to }{\mathop{l}}\,\times \overset{\to }{\mathop{R}}\,\]. It is represented by the right hand rule. The magnetic field inside the conductor is zero, while on the surface, the magnetic field is perpendicular to both the current density and the surface normal.
The Biot-Savart law is applicable only for steady currents.
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