A straight wire AB of mass 40 g and length 50 cm is suspended by a pair of flexible leads in uniform magnetic field of magnitude 0.40 T as shown in the figure. The magnitude of the current required in the wire to remove the tension in the supporting leads is ___________ A.
( Take g=10 ms⁻² ).

Solution

Physics sir · 4 steps

The leads go slack when the magnetic push upward exactly holds up the wire's weight.

1. Direction first. The field points into the page. With current flowing from A to B, F⃗ = IL⃗ × B⃗ points straight up, against gravity.
2. Zero tension. If the leads carry no tension, only two forces act on the wire: the magnetic force up and the weight down. They must be equal.

   BIL = mg ⇒ I = mg/BL

3. Put in numbers. In SI units, m = 0.04 kg and L = 0.5 m. So mg = 0.04 × 10 = 0.4 N and BL = 0.40 × 0.5 = 0.2 T m.
4. So. I = 0.4/0.2 = 2 A, flowing from A to B. The answer is 2.

The common slip: 2000. Leaving the mass in grams makes mg = 400 and gives 2000 A, not 2 A. Use m = 0.04 kg.

The idea to keep. To make supporting strings go slack, set the magnetic force BIL equal to mg. Convert grams to kilograms and centimetres to metres before you divide.

Answer from the NTA final answer key. Solution by PrepWiser.