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

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 ______ $μT$ . $(\frac{μ_{0}}{4 π} = 10^{- 7} N / A^{2})$ [2026]

Smart Achievers · Accepted Answer

(2)

$B_{net} = B_{1} + B_{2} = 2 \times \frac{μ_{0} i}{2 π r} = \frac{μ_{0} i}{π r}$

$B_{net} = \frac{4 π \times 10^{- 7} \times 30}{π \times 4 \times 10^{- 2}} = 30 \times 10^{- 5} T = 300 μ T$

Solution

Q.
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 ______ $μT$ . $(\frac{μ_{0}}{4 π} = 10^{- 7} N / A^{2})$ [2026]

1 30

2 300

3 150

4 0.1

Ans.
(2)

$B_{net} = B_{1} + B_{2} = 2 \times \frac{μ_{0} i}{2 π r} = \frac{μ_{0} i}{π r}$

$B_{net} = \frac{4 π \times 10^{- 7} \times 30}{π \times 4 \times 10^{- 2}} = 30 \times 10^{- 5} T = 300 μ T$

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