Torque On Current Carrying Coil In External Magnetic Field

Description:

Case 1:

When plane of the loop is perpendicular to magnetic field length of AB = DC = 𝑙 and that of BC = AC = 𝑏 .

Force experienced by all the sides are shown in the figure, force on AB and DC are equal and opposite to each other and that on BC and AD too.

Since the line of action of the forces on AB and DC is same and also the line of action of the forces BC and AD is same, therefore torque is zero.

Case 2:

When the plane of the loop is inclined to the magnetic field.

In this case again,

Lines of action of the forces on AB and DC are different, therefore this forms a couple and produces a torque. Side view of the loop is shown in the figure.

Torque = BI l(b sinθ) = BI (lb) sinθ

⇒ BIA sinθ

If loop has N turns then,

τ = BNIA sinθ

In vector form

τ = →μ → × B →

Where, μ = →NIA →

Energy needed to rotate the loop in uniform magnetic field

Energy needed to rotate the loop through an angle dθ is dU = τdθ

Where A→ is the area vector of the loop whose magnitude is area of the loop and direction is out of the plane for anti-clock wise sense of the current and into the loop for clockwise work of current.

Energy needed to rotate the loop through an angle dθ is dU = τdθ

ΔU = dU θ₂∫θ₁ τdθ = θ₂∫θ₁ μB sinθdθ

ΔU = μB(cosθ₁ − cosθ₂)

Energy stored in current carrying loop in uniform magnetic field

Then, energy stored in the loop is-

U = −μ_m.→B→