
Calculate the magnitude of the magnetic field at a point $100\text{ }cm$ from a long thin conductor carrying a current of $1\text{ }A$?
Answer
467.1k+ views
Hint: We have to find the magnetic field for a point which is at a distance $100\text{ }cm$ from the long wire. It is given that the wire is long and straight, so it can be treated as a long infinite wire. By substituting the values of the variables in the equation to find the magnetic field due to a long infinite wire, we will find the answer.
Complete step by step answer:
In the given question it is stated that we have to find the magnetic field of a point which is at a distance $100\text{ }cm$ from a long current carrying wire with a current of $1\text{ }A$.
According to Biot-Savart’s Law, the magnetic field at a point due to an wire is,
$B=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{I}{a}\left[ \sin {{\phi }_{1}}+\sin {{\phi }_{2}} \right]--\left( 1 \right)$
The variables are defined as,
$B=$ Magnetic field
$I=$ current through the wire
$a=$ distance between the point and the wire.
${{\phi }_{1}}$ and ${{\phi }_{2}}$ are the angles that the point makes with both the ends.
It is said in the question that the wire is long and straight. Hence, we can consider it as an infinite wire.
In case of infinite wire we get, ${{\phi }_{1}}={{\phi }_{2}}=\dfrac{\pi }{2}$
Hence, $\left[ \sin {{\phi }_{1}}+\sin {{\phi }_{2}} \right]=\sin \dfrac{\pi }{2}+\sin \dfrac{\pi }{2}=1+1=2$
Substituting the value in equation $\left( 1 \right)$ we get, the magnetic field due to a long infinite wire is,
$B=\dfrac{{{\mu }_{0}}}{2\pi }\dfrac{I}{a}--\left( 2 \right)$
Hence
The value of $\dfrac{{{\mu }_{0}}}{2\pi }=2\times {{10}^{-7}}$
Given,
$I=$ current through the wire$=1\text{ }A$
$a=$ the distance between the point and the wire$=100\text{ }cm=1\text{ }m$
Substituting the values in the equation $\left( 2 \right)$ we get,
$B=2\times {{10}^{-7}}\times \dfrac{1}{1}=2\times {{10}^{-7}}$
The magnetic field at a point which is at a distance $100\text{ }cm$ from a long thin conductor carrying a current of $1\text{ }A$ is $2\times {{10}^{-7}}\text{ }T$.
Note:
It must be noted that in case of Biot-Savart’s Law due to an infinite wire, the point makes angle approximately, ${{90}^{\circ }}$ with both the ends of the wire as the wire is infinite. The original formula for the Biot-Savart’s Law is $B=\dfrac{{{\mu }_{0}}}{2\pi }\dfrac{I}{a}\left[ \sin {{\phi }_{1}}+\sin {{\phi }_{2}} \right]$.
Complete step by step answer:
In the given question it is stated that we have to find the magnetic field of a point which is at a distance $100\text{ }cm$ from a long current carrying wire with a current of $1\text{ }A$.
According to Biot-Savart’s Law, the magnetic field at a point due to an wire is,
$B=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{I}{a}\left[ \sin {{\phi }_{1}}+\sin {{\phi }_{2}} \right]--\left( 1 \right)$
The variables are defined as,
$B=$ Magnetic field
$I=$ current through the wire
$a=$ distance between the point and the wire.
${{\phi }_{1}}$ and ${{\phi }_{2}}$ are the angles that the point makes with both the ends.
It is said in the question that the wire is long and straight. Hence, we can consider it as an infinite wire.
In case of infinite wire we get, ${{\phi }_{1}}={{\phi }_{2}}=\dfrac{\pi }{2}$
Hence, $\left[ \sin {{\phi }_{1}}+\sin {{\phi }_{2}} \right]=\sin \dfrac{\pi }{2}+\sin \dfrac{\pi }{2}=1+1=2$
Substituting the value in equation $\left( 1 \right)$ we get, the magnetic field due to a long infinite wire is,
$B=\dfrac{{{\mu }_{0}}}{2\pi }\dfrac{I}{a}--\left( 2 \right)$
Hence
The value of $\dfrac{{{\mu }_{0}}}{2\pi }=2\times {{10}^{-7}}$
Given,
$I=$ current through the wire$=1\text{ }A$
$a=$ the distance between the point and the wire$=100\text{ }cm=1\text{ }m$
Substituting the values in the equation $\left( 2 \right)$ we get,
$B=2\times {{10}^{-7}}\times \dfrac{1}{1}=2\times {{10}^{-7}}$
The magnetic field at a point which is at a distance $100\text{ }cm$ from a long thin conductor carrying a current of $1\text{ }A$ is $2\times {{10}^{-7}}\text{ }T$.
Note:
It must be noted that in case of Biot-Savart’s Law due to an infinite wire, the point makes angle approximately, ${{90}^{\circ }}$ with both the ends of the wire as the wire is infinite. The original formula for the Biot-Savart’s Law is $B=\dfrac{{{\mu }_{0}}}{2\pi }\dfrac{I}{a}\left[ \sin {{\phi }_{1}}+\sin {{\phi }_{2}} \right]$.
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