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Question: A small current element \({{Id\vec l}}\), with \({\text{d\vec l}} = 2\overset{\lower0.5em\hbox{\)\sm...

A small current element Idl⃗{{Id\vec l}}, with {\text{d\vec l}} = 2\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{k} {\text{ }}mm and I=2A{\text{I}} = 2{\text{A}} is centred at the origin. Find magnetic field dB⃗d\vec B āt the following points:
(A) On the x-axis at x=3mx = 3m
(B) On the x-axis at x=6mx = 6m
(C) On the z-axis at z=3mz = 3m.

Explanation

Solution

To solve this question, we need to use the vector form of the Biot Savart’s law. For this we need to substitute the position vectors of the points given in all the three parts of this question to find out the magnitude and direction of the magnetic field in the respective cases.
Formula used: The formula used to solve this question is given by
dB⃗=μ04πIdl⃗×r^r2d\vec B = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{Id\vec l \times \hat r}}{{{r^2}}}, here dB⃗d\vec B is the magnetic field produced by a current element Idl⃗Id\vec l at a point whose position vector is r⃗\vec r-with respect to the centre of the current element.

Complete step-by-step solution:
We know from the Biot Savart’s law that the magnetic field produced by a current element at a point is given by
dB⃗=μ04πIdl⃗×r^r2d\vec B = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{Id\vec l \times \hat r}}{{{r^2}}} ………………………..(1)
Now, according to the question, we have {\text{d\vec l}} = 2\hat k{\text{ }}mm = 2 \times {10^{ - 3}}\hat k{\text{ }}mm and I=2A{\text{I}} = 2{\text{A}}. Also we know that μ04π=10−7Tm/A\dfrac{{{\mu _0}}}{{4\pi }} = {10^{ - 7}}Tm/A. Substituting these in (1) we get
dB⃗=10−72×2×10−3(k^×r^)r2d\vec B = {10^{ - 7}}\dfrac{{2 \times 2 \times {{10}^{ - 3}}\left( {\hat k \times \hat r} \right)}}{{{r^2}}}
⇒dB⃗=4×10−10(k^×r^)r2\Rightarrow d\vec B = 4 \times {10^{ - 10}}\dfrac{{\left( {\hat k \times \hat r} \right)}}{{{r^2}}}............................(2)
We have to determine the magnetic field on the x-axis at x=3mx = 3m. The position vector of this point can be given by r⃗=3i^ m\vec r = 3\hat i{\text{ }}m. This means that r^=i^\hat r = \hat i and r=3mr = 3m. Substituting these in (2) we get
dB⃗=4×10−10(k^×i^)32d\vec B = 4 \times {10^{ - 10}}\dfrac{{\left( {\hat k \times \hat i} \right)}}{{{3^2}}}
On solving we get
dB⃗=4.44×10−11j^d\vec B = 4.44 \times {10^{ - 11}}\hat j
We have to determine the magnetic field on the x-axis at x=6mx = 6m. The position vector of this point can be given by r⃗=6i^ m\vec r = 6\hat i{\text{ }}m. This means that r^=i^\hat r = \hat i and r=6mr = 6m. Substituting these in (2) we get
dB⃗=4×10−10(k^×i^)62d\vec B = 4 \times {10^{ - 10}}\dfrac{{\left( {\hat k \times \hat i} \right)}}{{{6^2}}}
On solving we get
dB⃗=1.11×10−11j^d\vec B = 1.11 \times {10^{ - 11}}\hat j
We have to determine the magnetic field on the z-axis at z=3mz = 3m. The position vector of this point can be given by r⃗=3k^ m\vec r = 3\hat k{\text{ }}m. This means that r^=k^\hat r = \hat k and r=3mr = 3m. Substituting these in (2) we get
dB⃗=4×10−10(k^×k^)32d\vec B = 4 \times {10^{ - 10}}\dfrac{{\left( {\hat k \times \hat k} \right)}}{{{3^2}}}
On solving we get
dB⃗=0d\vec B = 0

Note: Prefer to use the vector form of the Biot Savart’s law to solve these types of questions. This is because otherwise we have to calculate the angle between the position vector and the current element, which will make the solution complex.