Physics · Moving Charges And Magnetism · NEET
The magnetic force must point straight up and its size must equal the weight. Magnetic force = B I L (when the wire is perpendicular to B). Weight = m g (downward). Setting them equal gives the balance condition B I L = m g. From this, the required current is I = m g / (B L). If the current is bigger than this, the wire rises; if smaller, it falls.
Use F = I L x B and the right-hand rule (or Fleming's left-hand rule). Point fingers along the current direction, curl toward B; the thumb gives the force. To suspend the wire, you must choose the current direction so this force is vertically UP, opposite to gravity. If your first current direction pushes the wire down, just reverse the current.
Both forms appear in NEET. If the whole wire has mass m and length L, use B I L = m g. If the problem gives mass per unit length (lambda, in kg/m), notice L cancels: B I L = (lambda·L) g, so B I = lambda g and I = lambda g / B. Read carefully whether the number is total mass or mass per metre.
For the cleanest balance, yes. Force is B I L sin(theta). It is maximum when the wire is perpendicular to B (theta = 90 degrees, sin = 1). If the wire is not perpendicular, only B I L sin(theta) is available to fight gravity, so you need a larger current: I = m g / (B L sin(theta)).
Yes, it is the reverse case. A wire lying on the floor is pulled UP by a parallel current above it (like currents attract) with force per length μ0·I1·I2/(2π·h). It stays on the floor as long as this upward pull does not exceed its weight per length λg. Setting them equal gives the just-lifts-off condition — this is exactly the ReNEET 2026 problem below.
A long straight wire of length 2 m and mass 250 g is suspended horizontally in a uniform horizontal magnetic field of 0.7 T. The amount of current flowing through the wire will be (g = 9.8 m/s^2):
Two infinitely long parallel wires A and B carry currents I and 2I in the same direction. Wire A has mass per unit length λ and lies on an insulated floor. Wire B is fixed at height h above the floor. The minimum h so that wire A does NOT rise from the floor is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
B I L = m g. Rearranged, the current needed is I = m g / (B L), where m = mass, L = length, B = magnetic field. The magnetic force B I L must point upward to cancel the weight.
The current in the field feels a magnetic force F = B I L. If you choose the current direction so this force points up and make it equal to m g, the net force is zero, so the wire hangs still (equilibrium).
Use B I L = λ L g. The length L cancels on both sides, leaving B I = λ g, so I = λ g / B. This is common in NEET numericals to test whether you noticed the units.
No, it is a delicate (neutral/unstable) balance. If the current or field changes slightly, the wire moves up or down. NEET usually asks only for the exact balancing current or field, not stability.
Yes. NCERT Class 12 Chapter 'Moving Charges and Magnetism' has the mid-air suspension example where m g = I L B gives B = m g /(I L) = 0.65 T for m = 0.2 kg, L = 1.5 m, I = 2 A.
