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A closely wound solenoid of 2000 turns and area of cross-section $1.6 \times 10^{-4} \mathrm{~m}^2$, carrying a current of $4.0$ $A$, is suspended through its centre allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of $7.5 \times 10^{-2} \mathrm{~T}$ is set up at an angle of $30^{\circ}$ with the axis of the solenoid?
(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of $7.5 \times 10^{-2} \mathrm{~T}$ is set up at an angle of $30^{\circ}$ with the axis of the solenoid?
Solution:
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Verified Answer
(a) Given: No. of turns $\mathrm{N}=2000$
area $=1.6 \times 10^{-4} \mathrm{~m}^2$
current $\mathrm{I}=4.0 \mathrm{~A}$, magnetic field
$\mathrm{B}=7.5 \times 10^{-2} \mathrm{~T}$
To find: Magnetic moment $\mathrm{M}$
Torque for $\theta=30^{\circ}$, Force.
Formula: $\mathrm{M}=\mathrm{NIA}$
$\tau=\mathrm{MB} \sin \theta$
The magnetic moment
$\mathrm{M}=\mathrm{NIA}=40 \times 1.6 \times 10^{-4} \times 200=1.28 \mathrm{Am}^2$
(b) Torque $\tau=\mathrm{MB} \sin \theta=1.28 \times 7.5 \times 10^{-2} \sin$
$$
30^{\circ}=0.096 \times \frac{1}{2}=0.048 \mathrm{Nm}
$$
Since force on both ends is equal and opposite hence net force is zero.
area $=1.6 \times 10^{-4} \mathrm{~m}^2$
current $\mathrm{I}=4.0 \mathrm{~A}$, magnetic field
$\mathrm{B}=7.5 \times 10^{-2} \mathrm{~T}$
To find: Magnetic moment $\mathrm{M}$
Torque for $\theta=30^{\circ}$, Force.
Formula: $\mathrm{M}=\mathrm{NIA}$
$\tau=\mathrm{MB} \sin \theta$
The magnetic moment
$\mathrm{M}=\mathrm{NIA}=40 \times 1.6 \times 10^{-4} \times 200=1.28 \mathrm{Am}^2$
(b) Torque $\tau=\mathrm{MB} \sin \theta=1.28 \times 7.5 \times 10^{-2} \sin$
$$
30^{\circ}=0.096 \times \frac{1}{2}=0.048 \mathrm{Nm}
$$
Since force on both ends is equal and opposite hence net force is zero.
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