Question

Two long straight parallel wires separated by 20 cm carry 5 A and 10 A current respectively, in the same direction. Find the magnitude and direction of the net magnetic field at a point midway between them.

Hint:

When calculating the magnetic field due to parallel currents, use Ampere's law and apply the right-hand rule to determine the direction of the field. The fields due to currents in the same direction will oppose each other.

Answer 1

Step 1: The magnetic field \( B \) produced by a long straight current-carrying conductor at a distance \( r \) is quantified by Ampere's law: \[ B = \frac{\mu_0 I}{2 \pi r} \] Here, \( B \) represents the magnetic field at distance \( r \) from the wire, \( I \) is the current flowing through the wire, \( r \) is the radial distance from the wire, and \( \mu_0 \) denotes the permeability of free space (\( \mu_0 = 4\pi \times 10^{-7} \, \text{}^2 \)).
Step 2: When two wires carry currents in the same direction, the magnetic field from each wire at a midpoint between them (where \( r = 10 \, \text{cm} = 0.1 \, \text{m} \) from both) is calculated using the aforementioned formula.
For the first wire with current \( I_1 = 5 \, \text{A} \):
\[ B_1 = \frac{\mu_0 I_1}{2 \pi r} = \frac{(4\pi \times 10^{-7}) \times 5}{2 \pi \times 0.1} = 1 \times 10^{-5} \, \text{T} \]
For the second wire with current \( I_2 = 10 \, \text{A} \):
\[ B_2 = \frac{\mu_0 I_2}{2 \pi r} = \frac{(4\pi \times 10^{-7}) \times 10}{2 \pi \times 0.1} = 2 \times 10^{-5} \, \text{T} \]
Step 3: Given that the currents are in the same direction, the magnetic fields generated by the two wires at the midway point will oppose each other. Consequently, the resultant magnetic field at this midpoint is the difference between these two fields:
\[ B_{\text{net}} = B_2 - B_1 = 2 \times 10^{-5} - 1 \times 10^{-5} = 1 \times 10^{-5} \, \text{T} \]
Step 4: The direction of the magnetic field is ascertained via the right-hand rule. The magnetic field from the first wire points into the page, while the magnetic field from the second wire points out of the page. Since these fields are in opposing directions, the net magnetic field is directed out of the page.
Therefore, the magnitude of the net magnetic field is \( 1 \times 10^{-5} \, \text{T} \), and its direction is out of the page.