Magnetic field due to currentMedium
Question
A current of 30 A each flows in opposite directions in two conducting wires, placed parallel to each other at a distance of 8 cm . The magnetic field at the mid point between the two wires is $\_\_\_\_$ $\mu T$.
$$\left( \frac{\mu_{0}}{4\pi} = 10^{- 7}\text{ }N/A^{2} \right) $$(1) 30
Options
A.30
B.150
C.0.1
Solution
Sol. [Key Concept: Opposite currents make the fields add at the midpoint: $B = 2 \times \frac{\mu_{0}I}{2\pi\ d}$ with $d = 4$ cm.]
$${B = \frac{\mu_{0}I}{2\pi(4) \times 10^{- 2}} \times 2 }{= 10^{- 7} \times 30 \times 10^{2} }{= 30 \times 10^{- 5}\text{ }T }{= 300\mu\text{ }T }$$
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