Question

A square coil of edge l and with n turns carries a current i. It is kept on a smooth horizontal plate. A uniform magnetic field B exists parallel to an edge. The total mass of the coil is M. What should be the minimum value of B for which the coil will start tipping over?

Answer 1

Explanation:

Turns of the coil is denoted as n

Square loop’s edge = l

Intensity of magnetic field = B

Current’s magnitude = i

Angle between the magnetic field and the area vector, θ = 90°

Due to magnetic field, the torque on the coil,

τ = niABsinθ

Here, the coil’s area is A.

τ = nil2B

Due to the weight, the torque produced, τweight = mgl2

The coil starts tipping over,

τ ≥ τweight

For the B minimum value,

τ = τweight

⇒nil2B = mgl2

⇒B = mg2nil