Physics 2007 JAMB Past Questions - Question 24: Two long parallel wires X and Y carry currents 3A and 5A respectively. If the force experienced per unit length by X is 5 x 10-N, the force per unit length experienced by wire Y is

Correct Answer

D

Explanation

since the two wires are parallel to each other the forces that they exerts on each other are equal and opposite, since they must agree with the new-ton's law of action and reaction. Hence the force experienced by the wire Y is equally 5x10-N
The force per unit length experienced by a wire due to another parallel wire carrying current can be calculated using Ampere's law. Ampere's law states that the magnetic field created by a long straight current-carrying wire at a distance 'r' from it is given by:

\[B = \frac{\mu_0 I}{2\pi r},\]

where:
- \(B\) is the magnetic field in Tesla (T).
- \(\mu_0\) is the permeability of free space, approximately equal to \(4\pi \times 10^{-7}\) T·m/A.
- \(I\) is the current in the wire in Amperes (A).
- \(2\pi r\) is the circumference of a circle with radius 'r' centered on the wire.

The force per unit length experienced by a wire carrying a current 'I_1' due to another wire carrying a current 'I_2' is given by:

\[F = \frac{\mu_0 I_1 I_2}{2\pi r},\]

Now, you are given that wire X carries a current of 3 A, and it experiences a force per unit length of \(5 \times 10^{-5}\) N. We can rearrange the formula above to find the force per unit length experienced by wire Y (carrying a current of 5 A):

\[F_Y = \frac{\mu_0 I_X I_Y}{2\pi r},\]

where
- \(I_X\) is the current in wire X (3 A).
- \(I_Y\) is the current in wire Y (5 A).
- \(\mu_0\) is the permeability of free space.

Substitute the known values:

\[F_Y = \frac{(4\pi \times 10^{-7}\, \text{T}\cdot\text{m}/\text{A})(3\, \text{A})(5\, \text{A})}{2\pi r} = \frac{60\times 10^{-7}\, \text{T}\cdot\text{m}/\text{A}}{2\pi r} = 30 \times 10^{-7}\, \text{T}\cdot\text{m}/\text{A} = 3 \times 10^{-6}\, \text{N/m}.\]

So, the force per unit length experienced by wire Y is \(3 \times 10^{-6}\) N/m.