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Standard XII

Physics

RMS Value of Current in Sinusoidal AC

Two long, par...

Question

Two long, parallel conductors carry currents $i_{1} = 3.0 A$ and $i_{2} = 3.0 A$ both directed into the page. Determine the magnitude of the resultant magnetic field at point P.

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Solution

$B_{1} = \frac{μ_{o}}{2 π} \frac{i_{1}}{r_{1}}$

$= \frac{(2 \times 10^{- 7}) (3)}{(5 \times 10^{- 2})} = 1.2 \times 10^{- 5} T$
$B_{2} = \frac{μ_{o}}{2 π} \frac{i_{2}}{r_{2}}$
$= \frac{(2 \times 10^{- 7}) (3)}{(12 \times 10^{- 2})} = 0.5 \times 10^{- 5} T$
$B_{1}$ and $B_{2}$ are mutually perpendicular at P.
$∴ B_{n e t} = \sqrt{B_{1}^{2} + B_{2}^{2}}$ $= 1.3 \times 10^{- 5} T$

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Two long, parallel conductors carry currents i1=3.0A and i2=3.0A both directed into the page. Determine the magnitude of the resultant magnetic field at point P.

B1=μo2πi1r1 =(2×10−7)(3)(5×10−2)=1.2×10−5TB2=μo2πi2r2 =(2×10−7)(3)(12×10−2)=0.5×10−5TB1 and B2 are mutually perpendicular at P.∴Bnet=√B21+B22=1.3×10−5T

Two long, parallel conductors carry currents $i_{1} = 3.0 A$ and $i_{2} = 3.0 A$ both directed into the page. Determine the magnitude of the resultant magnetic field at point P.