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प्रश्न
A long wire with a small current element of length 1 cm is placed at the origin and carries a current of 10 A along the X-axis. Find out the magnitude and direction of the magnetic field due to the element on the Y-axis at a distance 0.5 m from it.
उत्तर
`vec|dB| = (mu_0)/(4 pi) (I dl sin theta ) / (r^2) `
dl = Δx = 10-2 m , I = 10A r = 0.5 m = y, `mu_0 /4 pi = 10^(-7) (Tm)/A`
`theta = 90^circ ; sin theta = 1 `
`vec|dB| = (10^(-7) xx 10 xx 10^(-2))/(25 xx 10^(-2)) = 4 xx 10^(-8) T`
The direction of the field is in the + z-direction. This is so since,
`vec(dl) xx vec(r) = Delta x hat (i) xx y hat (j) = y Delta x ( hat(i) xx hat(j) ) = y Delta x hat (k)`
Hence the direction of magnetic field is towards positive z-axis.
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