Question 269: Magnetism and Matter

Question

A coil having 100 turns, area of $$5 	imes 10^{-3} \mathrm{~m}^2$$, carrying current of $$1 \mathrm{~mA}$$ is placed in uniform magnetic field of $$0.20 \mathrm{~T}$$ such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through $$90^{\circ}$$ is _________ $$\mu \mathrm{J}$$.

Accepted Answer

To find the work done in turning the coil through $$90^{\circ}$$, we first need to understand the concept of torque on a current-carrying loop in a magnetic field and how work done relates to the change in potential energy of the system. The potential energy (U) of a magnetic dipole in a magnetic field is given by:
$$U = - \vec{M} \cdot \vec{B}$$
where:
$\vec{M}$ is the magnetic moment of the coil, and
$\vec{B}$ is the magnetic field.
For a coil with $N$ turns, carrying current $I$, and with an area $A$, the magnetic moment $\vec{M}$ is defined as:
$$M = NI \cdot A$$
Given that the coil has $$100$$ turns, carries a current of $$1 \, \mathrm{mA} = 1 	imes 10^{-3} \, \mathrm{A}$$, and the area of the coil is $$5 	imes 10^{-3} \, \mathrm{m}^2$$, we can calculate its magnetic moment as follows:
$$M = 100 \cdot 1 	imes 10^{-3} \cdot 5 	imes 10^{-3} = 0.5 	imes 10^{-3} \, \mathrm{Am}^2$$
Since the coil is initially placed such that its plane is perpendicular to the magnetic field (i.e., the angle $ 	heta = 0^{\circ} $), and then it is turned through $90^{\circ}$, the initial and final angles ($	heta_i$ and $	heta_f$) of the coil with respect to the magnetic field are $0^{\circ}$ and $90^{\circ}$ respectively. This means the initial potential energy ($U_i$) and final potential energy ($U_f$) of the system are:
$$U_i = - M B \cos(	heta_i)$$
$$U_f = - M B \cos(	heta_f)$$
Given that $B = 0.20 \, \mathrm{T}$, $	heta_i = 0^{\circ} \, (\cos(0) = 1)$, and $	heta_f = 90^{\circ} \, (\cos(90^{\circ}) = 0)$, the potential energies are:
$$U_i = - 0.5 	imes 10^{-3} \cdot 0.20 \cdot 1 = - 1 	imes 10^{-4} \, \mathrm{J}$$
$$U_f = - 0.5 	imes 10^{-3} \cdot 0.20 \cdot 0 = 0 \, \mathrm{J}$$
The work done ($W$) is equal to the change in potential energy:
$$W = U_f - U_i$$
$$W = 0 - (- 1 	imes 10^{-4}) = 1 	imes 10^{-4} \, \mathrm{J}$$
Therefore, the work done in turning the coil through $90^{\circ}$ is $$100 \mu \mathrm{J}$$.

Magnetism and Matter

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