Name: Conversion of a Galvanometer Into a Voltmeter
Uploaded: 2019-07-19
Duration: 7 min 18 s
Description: This video helps us understand the working of a galvanometer. It also teaches us how to convert it into a voltmeter of the desired range both theoretically a...

Question

With the help of Biot-Savart law derive the expression for the intensity of magnetic field at a point due to an infinitely long straight conductor carrying current.

Accepted Answer

Consider a small current element $l m$ of length $d l$ . Let $I$ be the current flowing through it. Let $P$ be a point at a distance $r$ from the current element. Then the magnetic field $d B$ at the point $P$ is given by the Biot-Savart's Law.

According to Biot-Savart's law, the magnitude of the magnetic field $d B$ at point $P$ due to the current element is :

1. $d B$ is directly proportional to the current $I$ .
2. $d B$ is directly proportional to the length element $d l$ .
3. $d B$ is directly proportional to the sine of angle between $d l$ and $r$ .
4. $d B$ is inversely proportional to the square of the distance $r$ between the point and the current element.

$d B \propto \frac{I d l \sin θ}{r^{2}}$

$d B = \frac{μ_{0}}{4 π} \frac{I d l \sin θ}{r^{2}}$

Consider a straight finite conductor $X Y$ carrying a current $I$ . Let $P$ be a point at a perpendicular distance $d$ from the wire. Consider an arbitrary length element $\vec{d l}$ at $O$ . Let $\vec{r}$ be the position vector of the point $P$ with respect to $\vec{d l}$ . $θ$ is the angle between $\vec{d l}$ and $\vec{r}$ .

According to Biot-Savart's law, the magnetic field $d \vec{B}$ at point $P$ is

$d B = \frac{μ_{0}}{4 π} \frac{I d l \sin θ}{r^{2}}$ ................... $(1)$

In the right-angled triangle $P O C$ , $θ + ϕ = 90 ° \Rightarrow θ = 90 ° - ϕ$

$\sin θ = \sin (90 - ϕ) = \cos ϕ$ ............ (2)

$\cos ϕ = \frac{d}{r} \Rightarrow r = \frac{d}{\cos ϕ}$ .............. $(3)$

$\tan ϕ = \frac{l}{d} \Rightarrow l = d \tan ϕ$

Differentiating we get $d l = d \sec^{2} ϕ d ϕ$ .............. $(4)$

Substituting $(2), (3), (4)$ in equation in $(1)$

$d B = \frac{μ_{0}}{4 π} \frac{I (d \sec^{2} ϕ d ϕ) \cos ϕ}{(\frac{d^{2}}{\cos^{2} ϕ})} = \frac{μ_{0}}{4 π} \frac{I}{d} \cos ϕ d ϕ$

The total magnetic field is obtained by integrating the above equation within the limits $- ϕ_{1}$ and $ϕ_{2}$

$B = \int_{- ϕ_{1}}^{ϕ_{2}} d B = \int_{- ϕ_{1}}^{ϕ_{2}} \frac{μ_{0}}{4 π} \frac{I}{d} \cos ϕ d ϕ$

$B = \frac{μ_{0}}{4 π} \frac{I}{d} {[\sin ϕ]}_{- ϕ_{1}}^{ϕ_{2}}$

$B = \frac{μ_{0}}{4 π} \frac{I}{d} [\sin ϕ_{2} - \sin (- ϕ_{1})]$

$B = \frac{μ_{0}}{4 π} \frac{I}{d} [\sin ϕ_{2} + \sin ϕ_{1}]$

If the conductor $X Y$ is of infinite length, then $ϕ_{1} = ϕ_{2} = 90 °$

Then $B = \frac{μ_{0}}{4 π} \frac{I}{d} [\sin 90 ° + \sin 90 °] = \frac{μ_{0}}{4 π} \frac{I}{d} \times (2)$

The magnetic field at $P$ is $B = \frac{μ_{0} I}{2 π d}$

With the help of Biot-Savart law derive the expression for the intensity of magnetic field at a point due to an infinitely long straight conductor carrying current.

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