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

Obtain an expression for the magnetic induction at a point due to an infinitely long straight conductor carrying current.

Accepted Answer

Magnetic induction due to infinitely long straight conductor carrying current: XY, is an infinitely long straight conductor carrying a current I(Figure).

P is a point at a distance a from the conductor. AB is a small element of length dl. θ is the angle between the current element $Idl$ And the line joining the element dl and the point P. According to Biot-Savart law, the magnetic induction at the point P due to the current element $Idl$ Is.

$dB = \frac{μ_{o}}{4 π} \frac{Idl}{r^{2}} \cdot sinθ . . . . . . (1)$

AC is drawn perpendicular to BP from A.

$∠ OPA = ϕ, ∠ APB = dϕ$

In $ΔABC, sinθ = ABAC = dlAC$

$∴ AC = dlsinθ . . . . . . . . . . (2)$

From $ΔAPC, AC = rdϕ . . . . . . . . . (3)$

From equations (2) and (3),

$rdϕ = dlsinθ . . . . . . . . . . . . (4)$

Substituting equation (4) in equation (1)

$dB = \frac{μ_{o}}{4 π} \frac{Irdϕ}{r^{2}} = \frac{μ_{o}}{4 π} \frac{Idϕ}{ r} . . . . . . . . . . . . . (5)$

In $ΔOPA, cosϕ = \frac{a }{r}$

$∴ r = \frac{a}{cosϕ} . . . . . . . . . . (6)$

Substituting equation (6) in equation (5)

$dB = \frac{μ_{ο}}{4 π} \frac{1}{a} cosϕdϕ$

The total magnetic induction at P due to the conductor XY is

$B = \int_{- ϕ_{1}}^{ϕ_{2}} dB = \int_{- ϕ_{1}}^{ϕ_{2}} \frac{μ_{o}}{4 πa} I cosϕdϕ$

$B = \frac{μ_{o}}{4 πa} I [{sinϕ}_{1} + {sinϕ}_{2} ]$

For infinitely long conductor,

$ϕ_{1} = ϕ_{2} = 90^{o}$

$∴ B = \frac{μ_{o} I }{2 πa}$

If the conductor is places in a medium of permeability $μ_{0}$ , $B = \frac{μ_{o} I }{2 πa}$

Embibe Experts Solutions for Chapter: Magnetism And Magnetic Effects of Electric Current, Exercise 1: Exercise

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