A long wire with a small current element of length 1 cm is placed atthe origin and carries a current of 10A along X-axis. Find out themagnitude and direction of magnetic field due to the element on Y axisat a distance 0.5 m from it.
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The magnetic field on the element is B = 2.8×10^−4π.
Explanation:
Q 1 =Q 2 =45° (0.5/ 0.5 = tanθ)
B = M o / 4πr l / r (sinϕ 1 +sinϕ 2 )
= Mo /4π × 10/0.5 x 10^-2 × [ sin(45) + sin(45) ]
= 10^7 × x 10 / 10^−2 x 0.5 x √2
B = 2.8×10^−4π
Thus the magnetic field on the element is B = 2.8×10^−4π.
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