Question

(i) Derive the expression for the torque acting on a current carrying loop placed in a magnetic field.

(ii) Explain the significance of a radial magnetic field when a current carrying coil is kept in it.

Answer 1

(i) Torque on current carrying loop in a magnetic field

Consider a rectangular loop PQRS with sides PQ = l and PS = b. Let I be the current flowing in the loop placed in a uniform magnetic field \(\vec B\)  as shown in the figure.

Let θ be angle made by the plane of the coil with magnetic field \(\vec B\). Let \(\vec F_1, \ \vec F_2, \ \vec F_3\) and \(\vec F_4\) be forces acting on the arms PQ, QR, RS and SP respectively of the coil.

Force on arm SP

\(\vec F_4 = I(\vec b \times \vec B)\)

or F₄ = IbB sin (180° - θ)

or F₄ = BIb sin θ ..........(1) [upwards]

Force on arm QR

\(\vec F_2 = I(\vec b \times \vec B)\)

or F₂ = I bB sin θ

or F₂ = BIb sin θ ..........(2) [downwards]

Since the forces \(\vec F_4\) and \(\vec F_2\)  are equal in magnitude and acting in opposite direction along the same straight line, so they cancel each other's effect. Hence produce no torque i.e. resultant torque is zero.

Force on arm PQ

\(\vec F_1 = I(\vec b \times \vec B)\)

or F₁ = IlB sin 90°

or F₁ = BIl ............(3) (towards the reader)

Force on arm RS

\(\vec F_3 = I(\vec b \times \vec B)\)

or F₃ = IlB sin 90°

or F₃ = BIl ..........(4) (away from the reader)

These two parallel, equal and opposite forces acting along different lines of action, so they form a couple. The moment of the couple i.e. torque τ is given by

τ = Either force x ⊥ distance between the forces

= BIl (PT)

= BIl b cos θ

τ = BIA cos θ...........(5)

(where A = l x b is the area of the rectangular coil)

If the coil has n turns, then

τ = nBIA cos θ ...........(6)

If normal to the plane of the coil, makes an angle Φ with B, then θ = 90 - Φ.

So τ = nBIA cos (90 - Φ) = nBIA sin Φ.

(ii) A radial field is always perpendicular to the plane of the coil, hence moving coil galvanometer has linear scale.