A brief note on Magnets and the Magnetic Field of a Current . Read it before a test. Try each question at the end before you open its answer.
The big ideas A magnet has a north pole and a south pole . Like poles repel; unlike poles attract. Field lines run from north to south outside a magnet and back inside it: closed loops. They crowd together where the field is strong, and never cross. A current has a magnetic field (Oersted). Round a straight wire the lines are circles. Right-hand thumb rule: thumb along the current, the fingers curl the way the field goes. A coil of N N N turns gives N N N times the field of one loop. A solenoid has a strong, uniform field inside, like a bar magnet that can be switched off. A soft iron core makes an electromagnet : much stronger, and it loses its magnetism when the current stops. The facts at a glance Field of a long wire: rule: B = μ 0 I 2 π r = 2 × 10 − 7 I r B = \dfrac{\mu_0 I}{2\pi r} = 2 \times 10^{-7}\,\dfrac{I}{r} B = 2 π r μ 0 I = 2 × 1 0 − 7 r I ; unit: tesla (T) Turns in each metre: rule: n = N L n = \dfrac{N}{L} n = L N ; unit: per metre Field in a solenoid: rule: B = μ 0 n I B = \mu_0 n I B = μ 0 n I ; unit: tesla (T) The constant: rule: μ 0 = 4 π × 10 − 7 \mu_0 = 4\pi \times 10^{-7} μ 0 = 4 π × 1 0 − 7 ; unit: T m/A A worked example A long wire carries 11 A 11\ \text{A} 11 A . Find its field 4 cm 4\ \text{cm} 4 cm away. Then find the field inside a solenoid of 360 360 360 turns, 0.2 m 0.2\ \text{m} 0.2 m long, carrying 0.8 A 0.8\ \text{A} 0.8 A .
The wire. B = 2 × 10 − 7 T m/A × 11 A 0.04 m = 5.5 × 10 − 5 T = 55 μ T B = 2 \times 10^{-7}\ \text{T m/A} \times \dfrac{11\ \text{A}}{0.04\ \text{m}} = 5.5 \times 10^{-5}\ \text{T} = 55\ \mu\text{T} B = 2 × 1 0 − 7 T m/A × 0.04 m 11 A = 5.5 × 1 0 − 5 T = 55 μ T . Turns in each metre. n = 360 0.2 m = 1800 n = \dfrac{360}{0.2\ \text{m}} = 1800 n = 0.2 m 360 = 1800 turns per metre. The solenoid. B = 4 π × 10 − 7 T m/A × 1800 m − 1 × 0.8 A = 1.81 × 10 − 3 T = 1.81 mT B = 4\pi \times 10^{-7}\ \text{T m/A} \times 1800\ \text{m}^{-1} \times 0.8\ \text{A} = 1.81 \times 10^{-3}\ \text{T} = 1.81\ \text{mT} B = 4 π × 1 0 − 7 T m/A × 1800 m − 1 × 0.8 A = 1.81 × 1 0 − 3 T = 1.81 mT . Practice with new numbers
A program to try The program works out both fields of the worked example.
Read it, then run it. It repeats the worked example.
1. Double the distance from the wire. What happens to its field?
Mistakes to avoid Leaving the distance in centimetres in B = 2 × 10 − 7 I r B = 2 \times 10^{-7}\,\dfrac{I}{r} B = 2 × 1 0 − 7 r I . Use metres. Using all the turns as n n n . n n n is the number of turns in each metre. Saying field lines start and stop at the poles. They are closed loops. Using steel for an electromagnet. Steel stays magnetised. Check yourself A wire on the page carries a current up the page. Just to the right of the wire, does its field point into the page or out of it? Stuck? Show a hint Show the answerA wire's field is 60 μ T 60\ \mu\text{T} 60 μ T at 2 cm 2\ \text{cm} 2 cm . What is it at 8 cm 8\ \text{cm} 8 cm ? Stuck? Show a hint Show the answerA solenoid has 1250 1250 1250 turns in each metre and carries 2 A 2\ \text{A} 2 A . What is the field inside? Take μ 0 = 4 π × 10 − 7 T m/A \mu_0 = 4\pi \times 10^{-7}\ \text{T m/A} μ 0 = 4 π × 1 0 − 7 T m/A . Stuck? Show a hint Show the answerWhy does a scrapyard crane's electromagnet drop its load when the current is switched off? Stuck? Show a hint Show the answer