Question:
Find magnetic field at mid point between two parallel infinite wires carrying current of 30 Amp as shown in the figure.
Find magnetic field at mid point between two parallel infinite wires carrying current of 30 Amp as shown in the figure.
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The magnetic field due to two parallel currents in the same direction will add up at the midpoint, while if the currents were in opposite directions, they would cancel each other.
Updated On: Apr 7, 2026
- 0 µT
- 300 µT
- 150 µT
- 30 µT
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The Correct Option is B
Solution and Explanation
Let the distance between the two parallel wires be \( d = 8 \, \text{cm} = 0.08 \, \text{m} \), and the current through each wire is \( I = 30 \, \text{A} \).
The magnetic field due to an infinite straight wire is given by Ampere's Law:
\[
B = \frac{\mu_0 I}{2 \pi r}
\]
Where:
- \( B \) is the magnetic field
- \( \mu_0 = 4 \pi \times 10^{-7} \, \text{T·m/A} \) is the permeability of free space
- \( I \) is the current
- \( r \) is the distance from the wire to the point where the magnetic field is being calculated
Step 1: Magnetic field due to the first wire.
At the midpoint, the distance from each wire to the point is \( r = \frac{d}{2} = \frac{0.08}{2} = 0.04 \, \text{m} \). The magnetic field due to the first wire is:
\[ B_1 = \frac{\mu_0 I}{2 \pi r} = \frac{4 \pi \times 10^{-7} \times 30}{2 \pi \times 0.04} = \frac{120 \times 10^{-7}}{0.08} = 1.5 \times 10^{-4} \, \text{T} = 150 \, \mu\text{T} \]
Step 2: Magnetic field due to the second wire.
The magnetic field due to the second wire at the midpoint will have the same magnitude but opposite direction. Thus, the total magnetic field at the midpoint will be the sum of both fields:
\[ B_{\text{total}} = B_1 + B_2 = 150 \, \mu\text{T} + 150 \, \mu\text{T} = 300 \, \mu\text{T} \] Final Answer: 300 µT
- \( B \) is the magnetic field
- \( \mu_0 = 4 \pi \times 10^{-7} \, \text{T·m/A} \) is the permeability of free space
- \( I \) is the current
- \( r \) is the distance from the wire to the point where the magnetic field is being calculated
Step 1: Magnetic field due to the first wire.
At the midpoint, the distance from each wire to the point is \( r = \frac{d}{2} = \frac{0.08}{2} = 0.04 \, \text{m} \). The magnetic field due to the first wire is:
\[ B_1 = \frac{\mu_0 I}{2 \pi r} = \frac{4 \pi \times 10^{-7} \times 30}{2 \pi \times 0.04} = \frac{120 \times 10^{-7}}{0.08} = 1.5 \times 10^{-4} \, \text{T} = 150 \, \mu\text{T} \]
Step 2: Magnetic field due to the second wire.
The magnetic field due to the second wire at the midpoint will have the same magnitude but opposite direction. Thus, the total magnetic field at the midpoint will be the sum of both fields:
\[ B_{\text{total}} = B_1 + B_2 = 150 \, \mu\text{T} + 150 \, \mu\text{T} = 300 \, \mu\text{T} \] Final Answer: 300 µT
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