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Question: A long straight wire carries a current of 35A. What is the magnitude of the field B at a point 20cm ...

A long straight wire carries a current of 35A. What is the magnitude of the field B at a point 20cm from the wire?

Explanation

Solution

Hint: We will start by stating some general result and we will start by understanding what a magnetic field is. Then we will assume the angles in the figure and apply the formula for the magnetic field due to a conductor. Refer to the solution below.

Formula used: μ0I4πr(sinθ1+sinθ2)\dfrac{{{\mu _0}I}}{{4\pi r}}\left( {\sin {\theta _1} + \sin {\theta _2}} \right), BP=μ0I2πr{B_P} = \dfrac{{{\mu _0}I}}{{2\pi r}}.

Complete step-by-step solution -
Magnetic field is typically used as an area in which magnetism operates. This magnetism force is usually created by moving charge or other magnetic substance.
The value of current as per given in the question is- I=35AI = 35A
The value of distance as per given in the question is- r=20cm=0.2mr = 20cm = 0.2m
We will state a general result over here. Let’s assume a current carrying conductor (refer to the figure below).

Let's assume that the angle APD\angle APD is θ1{\theta _1} and the angle DPC\angle DPC is θ2{\theta _2}.
Now, the magnetic field at any point due to the finite conductor will always be-
μ0I4πr(sinθ1+sinθ2)\Rightarrow \dfrac{{{\mu _0}I}}{{4\pi r}}\left( {\sin {\theta _1} + \sin {\theta _2}} \right)
As we know for an infinite conductor-
θ1=θ2=90\Rightarrow {\theta _1} = {\theta _2} = 90^\circ
The magnetic field at point P due to the finite conductor will be-
BP=μ0I2πr\Rightarrow {B_P} = \dfrac{{{\mu _0}I}}{{2\pi r}}
We know that the value of μ0{\mu _0} is 4π×1074\pi \times {10^{ - 7}}. Substituting this value along with the other values in the above equation-
BP=μ0I2πr  BP=4π×107×352π×0.2  BP=2×106×352  BP=35×106T  \Rightarrow {B_P} = \dfrac{{{\mu _0}I}}{{2\pi r}} \\\ \\\ \Rightarrow {B_P} = \dfrac{{4\pi \times {{10}^{ - 7}} \times 35}}{{2\pi \times 0.2}} \\\ \\\ \Rightarrow {B_P} = \dfrac{{2 \times {{10}^{ - 6}} \times 35}}{2} \\\ \\\ \Rightarrow {B_P} = 35 \times {10^{ - 6}}T \\\

Note: The actual rate of flow of charge is generally defined. We now know that fixed charges produce an electric field commensurate with the scale of the charge. It's the same theory that traveling charges generate magnetic fields equal to the current, which creates a magnetic influence on the existing current carrying conductor surrounding them. For example, passage of electrons on the nuclear orbits, this magnetic field is usually due to the subatomic particles in the conductor.