Q.38. A current of 30 A each flows in opposite directions in two conducting wires, placed parallel t

Q: What is the correct answer for this exam Physics Q38?

The correct answer is C. Let the two wires be $W_1$ and $W_2$. Current $I_1 = I_2 = 30 A$. Distance between wires $d = 8 cm = 0.08 m$. The midpoint is at $r_1 = r_2 = d/2 = 4 cm = 0.04 m$ from each wire. The magnetic field due to a long straight wire is $B = \frac{\mu_0 I}{2\pi r}$. Since the currents are in opposite directions, at the midpoint, the magnetic fields due to both wires will be in the same direction (e.g., if

Q: Q.38. 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…

The correct answer is C. Let the two wires be $W_1$ and $W_2$. Current $I_1 = I_2 = 30 A$. Distance between wires $d = 8 cm = 0.08 m$. The midpoint is at $r_1 = r_2 = d/2 = 4 cm = 0.04 m$ from each wire. The magnetic field due to a long straight wire is $B = \frac{\mu_0 I}{2\pi r}$. Since the currents are in opposite directions, at the midpoint, the magnetic fields due to both wires will be in the same direction (e.g., if

Q: What are the options for this this exam Physics question?

A. 30 · B. 300 · C. 150 · D. 1

Question

Q.38. 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 _______ . [Figure: Two parallel wires]

FirstInTest · Accepted Answer

Let the two wires be $W_1$ and $W_2$. Current $I_1 = I_2 = 30 A$. Distance between wires $d = 8 cm = 0.08 m$. The midpoint is at $r_1 = r_2 = d/2 = 4 cm = 0.04 m$ from each wire. The magnetic field due to a long straight wire is $B = \frac{\mu_0 I}{2\pi r}$. Since the currents are in opposite directions, at the midpoint, the magnetic fields due to both wires will be in the same direction (e.g., if one current is up and the other down, using the right-hand rule, both fields at the midpoint will point into or out of the page). So the total magnetic field $B_{total} = B_1 + B_2$. $B_{total} = \frac{\mu_0 I}{2\pi r} + \frac{\mu_0 I}{2\pi r} = \frac{2 \mu_0 I}{2\pi r} = \frac{\mu_0 I}{\pi r}$. Given $\mu_0 = 4\pi 	imes 10^{-7} T \cdot m/A$. $B_{total} = \frac{4\pi 	imes 10^{-7} 	imes 30}{\pi 	imes 0.04} = \frac{4 	imes 10^{-7} 	imes 30}{0.04} = \frac{120 	imes 10^{-7}}{0.04} = 3000 	imes 10^{-7} T = 300 	imes 10^{-6} T = 300 \mu T$.

Answer

30

Answer

300

Answer

150

Answer

1

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