First slide

Magnetic field due to current element

Question

Five very long, straight insulated wires are closely bound together to form a small cable. Currents carried by the wires are: $I_{1} = 20 A, I_{2} = - 6 A, I_{3} = 12 A, I_{4} = - 7 A$ , $I_{5} = 18 A$ .(Negative currents are opposite in direction to the positive.) The magnetic field induction at a distance of 10 cm from the cable is

Moderate

A

$5 μ T$
B

$15 μ T$
C

$74 μ T$
D

$128 μ T$

Solution

Net current is $(20 - 6 + 12 - 7 + 18) A$ , i.e., 37 A.
$r = \frac{10}{100} m = \frac{1}{10} m$
$B = \frac{μ_{0} I}{2 π r} = \frac{4 π \times 10^{- 7} \times 37 \times 10}{2 π \times 1} T = 74 \times 10^{- 6} T$
$= 74 μ T$ .

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