Question:

20 amp current is flowing in a long straight wire. The intensity of magnetic field at a distance of 10 cm from the wire, will be:

Updated On: Oct 10, 2023
  • $ 4\times {{10}^{-5}}wb/{{m}^{2}} $
  • $ 2\times {{10}^{-5}}wb/{{m}^{2}} $
  • $ 3\times {{10}^{-5}}wb/{{m}^{2}} $
  • $ 8\times {{10}^{-5}}wb/{{m}^{2}} $
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The Correct Option is A

Solution and Explanation

Here: $ ~i=20 $ amp, $ r=10\text{ }cm=10\times {{10}^{-2}}m $ Intensity of magnetic field produced due to straight current carrying wire will be $ B=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2i}{r} $ $ =\frac{{{10}^{-7}}\times 2\times 20}{0.1}=4\times {{10}^{-5}}\,wb/{{m}^{2}} $
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Concepts Used:

Amperes circuital law

According to Ampere’s law, magnetic fields are related to the electric current that is produced in them. This law specifies that the magnetic field is associated with a given current or vice-versa, provided that the electric field doesn’t change with time.

Ampere’s circuital law can be written as the line integral of the magnetic field surrounding the closed loop which is equal to the number of times the algebraic sum of currents passing through the loop.

According to the second equation, if the magnetic field is integrated along the blue path, then it is equal to the current enclosed, I.

The magnetic field doesn’t vary at a distance r because of symmetry. The path length (in blue) in figure 1 has to be equal to the circumference of a circle,2πr.