Magnetic fields due to currents — practice questions

Identify known values and convert units: Current $I = 12.0 \text{ A}$ Distance $r = 4.0 \text{ cm} = 0.040 \text{ m}$ Permeability of free space $\mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1}$

Choose the correct formula: The formula for magnetic flux density near a long straight wire is: $B = \frac{\mu_0 I}{2\pi r}$

Substitute the values into the formula: $B = \frac{(4\pi \times 10^{-7}) \times (12.0)}{2\pi \times (0.040)}$

Simplify and perform the calculation: Notice that $4\pi$ in the numerator and $2\pi$ in the denominator simplify to 2. $B = \frac{2 \times 10^{-7} \times 12.0}{0.040}$ $B = \frac{24.0 \times 10^{-7}}{0.040} = 600 \times 10^{-7} = 6.0 \times 10^{-5} \text{ T}$

Answer: The magnetic flux density at 4.0 cm from the cable is $6.0 \times 10^{-5} \text{ T}$.

Identify known values: Current $I_1 = 3.0 \text{ A}$ Current $I_2 = 5.0 \text{ A}$ Separation $r = 0.15 \text{ m}$ Permeability of free space $\mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1}$

Choose the correct formula: The formula for force per unit length between parallel wires is: $\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}$

Substitute the values into the formula: $\frac{F}{L} = \frac{(4\pi \times 10^{-7}) \times (3.0) \times (5.0)}{2\pi \times (0.15)}$

Perform the calculation: $\frac{F}{L} = \frac{2 \times 10^{-7} \times 15}{0.15} = \frac{30 \times 10^{-7}}{0.15} = 200 \times 10^{-7} = 2.0 \times 10^{-5} \text{ N m}^{-1}$

State the direction of the force: Since the currents are in the same direction, the force between the wires is attractive.

Answer: The force per unit length between the wires is $2.0 \times 10^{-5} \text{ N m}^{-1}$ (attractive).