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A circular coil of 20 turns and radius $10 \mathrm{~cm}$ is placed in a uniform magnetic field of $0.10 \mathrm{~T}$ normal to the plane of the coil. If the current in the coil is $5.0 \mathrm{~A}$, what is the
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross-sectional area $10^{-5} \mathrm{~m}^2$ and the free electron density in copper is given to about $10^{29} \mathrm{~m}^{-3}$.)
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross-sectional area $10^{-5} \mathrm{~m}^2$ and the free electron density in copper is given to about $10^{29} \mathrm{~m}^{-3}$.)
Solution:
2306 Upvotes
Verified Answer
The magnetic field is normal to the plane of the coil, so condition of minimum torque.
(a) Torque on the coil
$$
\tau=\mathrm{NI}_{\mathrm{AB}} \sin \theta
$$
Here, $\theta=0^{\circ} \therefore \tau=0$
(b) Force on every element of the coil cancelled force on corresponding element. Therefore, net force on the coil is zero.
(c) To calculate force on each electron, let us find drift velocity.
$I=$ Anev
$$
5=10^{-5} \times 10^{29} \times 1.6 \times 10^{-19} v_d
$$
or $v_d=3.125 \times 10^{-5} \mathrm{~ms}^{-1}$ Now force, $F=e v_d B$
$$
F=1.6 \times 10^{-19} \times 3.125 \times 10^{-5} \times 0.1=5 \times 10^{-25} \mathrm{~N}
$$
(a) Torque on the coil
$$
\tau=\mathrm{NI}_{\mathrm{AB}} \sin \theta
$$
Here, $\theta=0^{\circ} \therefore \tau=0$
(b) Force on every element of the coil cancelled force on corresponding element. Therefore, net force on the coil is zero.
(c) To calculate force on each electron, let us find drift velocity.
$I=$ Anev
$$
5=10^{-5} \times 10^{29} \times 1.6 \times 10^{-19} v_d
$$
or $v_d=3.125 \times 10^{-5} \mathrm{~ms}^{-1}$ Now force, $F=e v_d B$
$$
F=1.6 \times 10^{-19} \times 3.125 \times 10^{-5} \times 0.1=5 \times 10^{-25} \mathrm{~N}
$$
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