Name: Factors Affecting Magnetic Field of Straight Wire
Uploaded: 2019-07-19
Duration: 4 min 9 s
Description: We know a magnetic field is produced by current flowing in a straight conductor. In this video, let us learn the different factors affecting the intensity of...

Question

Derive an expression for torque on the rectangular coil of area $A$ carrying a current $I$ and placed in a uniform magnetic field $B$ . Indicate the direction of the torque on the loop.

Accepted Answer

Given the area of the coil $A$ , Current in the coil= $I$ , the external magnetic field= $B$ .

Consider a rectangular loop $PQRS$ of length $P Q = R S = l$ and breadth $QR = SP = b$ suspended in a uniform magnetic field $B$ . Let at any instant the normal to the plane of loop make an angle $θ$ with the direction of the magnetic field. Let $I$ be the current in the loop.

We know that a force acts on a current-carrying wire placed in a magnetic field. Therefore, each side of the loop will experience a force. Suppose that the forces on sides are $PQ = F_{1}, QR = F_{2}, RS = F_{3}, SP = F_{4}$ .
The sides QR and SP make an angle $(90 - θ)$ with the direction of the magnetic field. Therefore each of the forces $F_{2} and F_{4}$ acting on these sides has the same magnitude

$F = Blb \sin (90 - θ) = Blbcos θ$ ........................................................(1)
According to Flemings' left hand rule, the forces are equal and opposite, but their line of action is the same. Therefore, these cancel each other. The sides $PR ans RS$ of the current loop are perpendicular to the magnetic field, therefore the magnitude of each of the forces $F_{1} and F_{3}$ acting on sides $PQ and RS$ are equal and opposite, but their lines of action are different, therefore these forces form a couple called the defecting couple. When the normal to the plane of the loop makes an angle with the direction of the magnetic field the perpendicular distance between $F_{1} and F_{3}$ is $b sinθ$ .
Moment of couple or Torque is given by,

$T = magnitude of one force \times perpendicular distance$ .

$BIl \times B sinθ = (lb) B sinθ$ but $l \times b = A$ area of the loop, Hence the torque,

$T = IAB sinθ$ ., If the number of turns are $N$ , then the torque is,

$T = NIAB sinθ$ .............................................................(2)

In vector form, we can write,

$T = NI \vec{A} \times \vec{B}$

Since $N \times I \times A = M$ the magnet moment of the current loop, we get,

$T = \vec{M} \times \vec{B}$

This is the required expression.

Derive an expression for torque on the rectangular coil of area $A$ carrying a current $I$ and placed in a uniform magnetic field $B$ . Indicate the direction of the torque on the loop.

Important Questions on Electromagnetism

Learn and Prepare for any exam you want !

Derive an expression for torque on the rectangular coil of area A carrying a current I and placed in a uniform magnetic field B. Indicate the direction of the torque on the loop.

Important Questions on Electromagnetism

Learn and Prepare for any exam you want !

Derive an expression for torque on the rectangular coil of area $A$ carrying a current $I$ and placed in a uniform magnetic field $B$ . Indicate the direction of the torque on the loop.