Derive the expression for the torque on a rectangular coil of area A, carrying a current I, placed in a magnetic field B. The angle between the direction of B and the vector perpendicular to the plane of the coil is 0.

Solution

Let, a rectangular loop carrying a steady current I, placed in a uniform magnetic field B experiences a torque.
Angle between the field and the normal to the coil be angle $θ$ .

Force on arm BC and DA are equal and opposite, and is acting along the axis of the coil. Being, collinear along the axis they cancel each other resulting in no net force or torque.
Let, the forces on arms AB and CD be F₁ and F₂.
Magnitude of F₁ and F₂ is given by,
F₁= F₂ = BbI .
Since, these forces are equal and opposite net force is 0.
But, there is a torque acting on the loop as a result of these forces which, is given by

$τ = F_{1} \frac{a}{2} \sin θ + F_{2} \frac{a}{2} \sin θ = I ab B \sin θ = I A B \sin θ$

Now, as theta tends to 0, perpendicular distance between the force of the coule also approaches 0. This makes the forces collinear and, net force and torque becomes 0.

Therefore,
$τ = 0 for, θ \to 0$ .

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M₁ and M₂ have the following particulars:
R₁ = 10 Ω N₁ = 30,
A₁ = 3.6 x 10^–3 m², B₁ = 0.25 T
R₂ = 14 Ω; N₂ = 42,
A₂ = 1.8 x 10^–3 m², B₂ = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M_1.

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Moving Charges And Magnetism

Derive the expression for the torque on a rectangular coil of area A, carrying a current I, placed in a magnetic field B. The angle between the direction of B and the vector perpendicular to the plane of the coil is 0.

Some More Questions From Moving Charges And Magnetism Chapter

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