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Question: An element \[\overrightarrow{dl}=dx\hat{i}\](where \(dx\)=1cm) is placed at origin and carries a cur...

An element dl→=dxi^\overrightarrow{dl}=dx\hat{i}(where dxdx=1cm) is placed at origin and carries a current 10A. What is the magnetic field on y- axis at a dist. of 0.5m?
(A) 2×10−8k^T2\times {{10}^{-8}}\hat{k}T
(B) 4×10−8k^T4\times {{10}^{-8}}\hat{k}T
(C) −2×10−8k^T-2\times {{10}^{-8}}\hat{k}T
(D) −4×10−8k^T-4\times {{10}^{-8}}\hat{k}T

Explanation

Solution

We’re provided an element that produces a stable electric current of 10A. We know that such an element is capable of producing a magnetic field. The properties of this generated field is best explained by Biot-Savart law, hence, we use the same law to find the answer for this question.

Formulas used:
Biot Savart Law: B⃗=μ04πIdl⃗r⃗r3\vec{B}=\dfrac{\mu_0}{4\pi }\dfrac{Id\vec{l}\vec{r}}{{{r}^{3}}}, where B⃗\vec{B}the magnetic field due to the element dl⃗d\vec{l}of a wire that carries current II. rr is the distance between the element and the point, and r⃗\vec{r} is a unit vector that points from dl⃗d\vec{l}to our point.
μ0\mu_0 is the permeability of free space and its exact value is μ0=4π×10−7T⋅m/A\mu_0=4\pi \times {{10}^{-7}}T\cdot m/A

Complete step by step answer:
This question refers to the magnetic field on the Y-axis which depends on the distance of Y-axis
We’re given in the question that,
dl→=dxi^= 10−2m\overrightarrow{dl}=dx\hat{i}=\text{ }{{10}^{-2}}m
I=10AI = 10A
r = 0.5 mr\text{ }=\text{ }0.5\text{ }m
From Biot-Savart’s law,
B⃗=μ04πIdl⃗×r⃗r3\vec{B}=\dfrac{\mu_0}{4\pi }\dfrac{Id\vec{l}\times \vec{r}}{{{r}^{3}}}
⇒dl⃗×r⃗=dxi^+dyj⌢\Rightarrow d\vec{l}\times \vec{r}=dx\hat{i}+dy\overset{\scriptscriptstyle\frown}{j}
=ydx(i^+j⌢)=ydx(\hat{i}+\overset{\scriptscriptstyle\frown}{j})
=ydx(k⌢)=ydx(\overset{\scriptscriptstyle\frown}{k})
Hence, the direction of B⃗\vec{B}is in the z-direction.
dB=μ04πIdlsin⁡θr2dB=\dfrac{\mu_0}{4\pi }\dfrac{Idl\sin \theta }{{{r}^{2}}}
=4π×10−74π×10×1×10−2×sin⁡90∘0.5×0.5T=\dfrac{4\pi \times {{10}^{-7}}}{4\pi }\times \dfrac{10\times 1\times {{10}^{-2}}\times \sin {{90}^{\circ }}}{0.5\times 0.5}T
(θ=90∘\theta ={{90}^{\circ }}, since the axis of the element and y-axis are perpendicular to each other )
Hence, dB=4×10−8TdB=4\times {{10}^{-8}}T
The correct answer among the four is option B.

Note: Such an approximation in Biot-Savart law is feasible only when the length of the line segment is very small compared to the distance from the current element to the point. Else, the integral form of Biot-Savart law is to be used over the entire line segment to calculate the magnetic field.