A long straight conductor carries a current of 5 A. The magnitude of the magnetic field at a point 20 cm from the conductor is:

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  1. 5 μT
  2. 20 μT
  3. 10 μT
  4. 15 μT

Answer (Detailed Solution Below)

Option 1 : 5 μT
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Detailed Solution

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Concept:

  • A magnetic field has both magnitude and direction. Hence, it is a vector quantity denoted by B.
  • A magnetic field is a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials.
  • A moving charge in a magnetic field experiences a force perpendicular to its own velocity and to the magnetic field.
  • The magnitude of the magnetic field at a given point from the conductor is given as, \(B=\frac{μ_0}{4\pi}\frac{2i}{r}\)
  • Here, \(\frac{\mu_0}{4\pi}=10^{-7} N/m\)

Calculation:

Given,

The current is carried by the long straight conductor, i = 5 A

The distance of a point from a wire, r = 20 cm = 0.2 m

The magnitude of the magnetic field at a given point from the conductor is given as,

\(B=\frac{μ_0}{4\pi}\frac{2i}{r}\)

\(B=10^{-7}\times \frac{2\times 5}{0.2}=5μ T\)

Hence, the magnitude of the magnetic field at a point from the conductor is 5 μT.

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