Question
- Derive the expression for the torque acting on a current carrying loop placed in a magnetic field.
- Explain the significance of a radial magnetic field when a current carrying coil is kept in it.
✓
Answer
- The plane of the loop is not along the magnetic field but makes an angle with it.
Let the dimension of the rectangular coil $\text{ABCD},$ be $AB \times BC = a \times b$
The angle between the field and the normal is $\theta $.
Forces on $BC$ and $DA$ are equal and opposite and they cancel each other as they are collinear.
Force on $AB$ is $F_1$ and force on $CD$ is $F_2.$
$F_1 = F_2 = IbB$
The magnitude of the torque on the loop as in the figure:
$\therefore\tau=\text{F}_1\frac{\text{a}}{2}\sin\theta+\text{F}_2\frac{\text{a}}{2}\sin\theta$
$=\text{lab}\text{ B}\sin\theta$
$\tau=\text{lab}\sin\theta$
If there are $'n\ '$ such turns the torque will be $\text{n}\text{ lab}\sin\theta$
The magnetic moment of the current, $m = lA$
$\therefore\overrightarrow{\tau}=\overrightarrow{\text{m}}\times\overrightarrow{\text{B}}$
- The uniform radial magnetic field keeps the plane of the coil always parallel to the direction of the magnetic field. that is, the angle between the plane of the coil and the magnetic field is zero in all the orientation of the coil.
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