Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5 A. (see figure) (\mu _{0}=4\pi \times 10^{-7}N-A^{-2})

  • Option 1)

    2.0\times 10^{-5}T

     

     

     

  • Option 2)

    1.5\times 10^{-5}T

  • Option 3)

    3.0\times 10^{-5}T

  • Option 4)

    2.5\times 10^{-5}T

 

Answers (1)

 

Magnetic Field Due to a Straight Wire -

B=\frac{\mu_{o}i}{4\pi r}(\sin\phi_{1}+\sin\phi_{2})

- wherein

 

 

\sin \Theta =\frac{3}{5}

So B_{p}=\frac{\mu _{0}I}{4\pi d}\times 2\sin \Theta

B_{p}=\frac{10^{7}\times 4\pi }{4\pi }\times \frac{5\times 2}{4\times 10^{-2}}\times \frac{3}{5}

B_{p}=1.5\times 10^{-5}T

 


Option 1)

2.0\times 10^{-5}T

 

 

 

Option 2)

1.5\times 10^{-5}T

Option 3)

3.0\times 10^{-5}T

Option 4)

2.5\times 10^{-5}T

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