12.3 Magnetism and Current-Carrying Wires

1. Example 1Calculator allowed

   Field near a wire

   A long straight wire lies along the page and carries 10 A to the right. Find the size of the magnetic field 5.0 cm above the wire (in the plane of the page), and its direction there.

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   1. 1Step 1: $B = \dfrac{\mu_0 I}{2\pi r} = \dfrac{(4\pi \times 10^{-7})(10)}{2\pi(0.050)} = 4.0 \times 10^{-5}$ T.
   2. 2Step 2: Direction: right thumb along the current (to the right). Above the wire, your fingers come out of the page toward you.
   3. 3Step 3: That's close to the size of Earth's field, which is why a compass held near a wire with a few amps flowing visibly turns.

   Answer: 4.0 × 10⁻⁵ T, out of the page

2. Example 2Calculator allowed

   Two wires with opposite currents

   Two long parallel wires are 0.10 m apart. One carries 10 A and the other carries 20 A in the opposite direction. (a) Find the magnetic field at the midpoint between them. (b) Find the force on a 2.0 m length of each wire, and say whether they attract or repel.

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   1. 1Step 1: (a) Each wire is 0.050 m from the midpoint. The 10 A wire makes 4.0 × 10⁻⁵ T there and the 20 A wire makes 8.0 × 10⁻⁵ T. With opposite currents, both fields point the same way between the wires, so they add: 1.2 × 10⁻⁴ T.
   2. 2Step 2: (b) The field of the 10 A wire at the 20 A wire, 0.10 m away: $B = \dfrac{(4\pi \times 10^{-7})(10)}{2\pi(0.10)} = 2.0 \times 10^{-5}$ T.
   3. 3Step 3: Force on 2.0 m of the 20 A wire: F = IℓB = (20)(2.0)(2.0 × 10⁻⁵) = 8.0 × 10⁻⁴ N.
   4. 4Step 4: By Newton's third law, the 10 A wire feels the same 8.0 × 10⁻⁴ N in the opposite direction. Opposite currents repel.

   Answer: (a) 1.2 × 10⁻⁴ T. (b) 8.0 × 10⁻⁴ N on each wire, repelling.

3. Example 3Calculator allowed

   Wire between magnet poles

   A horizontal wire carries 5.0 A toward the east. A 0.30 m length of it sits in a uniform 0.40 T magnetic field pointing north. Find the size and direction of the force on the wire. What if the current flowed north instead?

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   1. 1Step 1: The current (east) is perpendicular to the field (north), so θ = 90°: F = IℓB = (5.0)(0.30)(0.40) = 0.60 N.
   2. 2Step 2: Direction: fingers east, curl toward north, thumb points straight up. The force is upward.
   3. 3Step 3: If the current flowed north, it would run along the field (θ = 0°), and the force would be zero.

   Answer: 0.60 N upward; zero if the current runs north

Common mistakes

• !Using 1/r² for a wire's field. It falls off as 1/r.
• !Mixing up the parallel-wire rule with the charge rule. Same-direction currents attract; same-sign charges repel.
• !Thinking the wire with more current feels the bigger force. The forces are a third-law pair and are equal.
• !Drawing a wire's field pointing toward or away from the wire. It circles around the wire.

On the exam

• ✓Expect to draw field directions around wires (into or out of the page) and to add the fields of two wires at a point.
• ✓Factor-of-change questions on B = μ₀I/(2πr) and F = IℓB are common, as is explaining why parallel wires attract or repel using one wire's field and the force on the other.