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Question: A copper wire of diameter \(1.6\,{\text{mm}}\) carries a current of \(20\,{\text{A}}\). Find the max...

A copper wire of diameter 1.6 mm1.6\,{\text{mm}} carries a current of 20 A20\,{\text{A}}. Find the maximum magnitude of the magnetic field B{\text{B}} due to this current.

Explanation

Solution

As you know current carrying wire produces a magnetic field around it. We can find the strength of this magnetic field with the most useful Biot-Savart law. Biot-Savart law states how the value of the magnetic field at a specific point in space from one short segment of current-carrying conductor depends on each factor that influences the field.

Complete step by step answer:
By using the Biot-Savart law
B = μ0NI2R{\text{B = }}\dfrac{{{\mu _0}{\text{NI}}}}{{2R}}
We can find a formula for current carrying straight wire to find the maximum magnetic strength, that is
B = μ0i2πr    ......(i){\text{B = }}\dfrac{{{\mu _0}{\text{i}}}}{{2\pi r}}\,\,\,\,......(i)
Now, according to question,
d = 1.6mm  =  1.6×10−3md\, = \,1.6{\text{mm}}\,{\text{ = }}\,{\text{1}}{\text{.6}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{m}}

Therefore, r = 1.62mm  =  0.8×10−3mr\, = \,\dfrac{{1.6}}{2}{\text{mm}}\,{\text{ = }}\,0.8 \times {\text{1}}{{\text{0}}^{ - 3}}{\text{m}}
Now by putting this value in the equation (i)(i) , we get
B = μ0× 202×π×0.8×10−3    ......(ii){\text{B = }}\dfrac{{{\mu _0} \times \,20}}{{2 \times \pi \times 0.8 \times {{10}^{ - 3}}}}\,\,\,\,......(ii)
Now put value of constant μ0= 4π× 10−7 N/A2{\mu _0} = \,4\pi \times \,{10^{ - 7}}\,{\text{N/}}{{\text{A}}^2} , equation (ii)(ii) becomes
B = 4π× 10−7× 202×π×0.8×10−3    ......(iii){\text{B = }}\dfrac{{4\pi \times \,{{10}^{ - 7}} \times \,20}}{{2 \times \pi \times 0.8 \times {{10}^{ - 3}}}}\,\,\,\,......(iii)
∴B = 0.005 T\therefore {\text{B = 0}}{\text{.005}}\,{\text{T}}

Hence, the maximum magnitude of the magnetic field B{\text{B}} due to the give current is 0.005 T{\text{0}}{\text{.005}}\,{\text{T}}.

Note: Biot-Savart law is similar to Coulomb's law in electrostatics.The law is applicable for very small conductors too which carry current.The law is applicable for symmetrical current distribution.We can use Biot–Savart law to calculate magnetic responses even at the atomic or molecular level.It is also used in aerodynamic theory to calculate the velocity induced by vortex lines.