Question

Query : A closely wound solenoid of 1000 turns and area of cross-section 2 × [tex] 10^{-4}
[/tex] m², carrying a current of 2.0 A, is placed with it's horizontal axis at 30° with the direction of a uniform horizontal magenitic field of 0.16 T.
a. What is the torque experienced by the solenoid due to the field?
b. If the solenoid is free to turn about the verticle direction, specify its orientations of stable and unstable equilibriums. What is the amount of work needed to displace the solenoid from its stable orientation to its unstable orientation?
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Answer 1

$\underline{\underline{\maltese\:\:\textbf{\textsf{Question}}}}$

A closely wound solenoid of 1000 turns and area of cross-section $\sf{2 \times 10^{-4} }$ m², carrying a current of 2.0 A, is placed with it's horizontal axis at 30° with the direction of a uniform horizontal magnetic field of 0.16 T

$\rule{345}{1}$

• a) What is the torque experienced by the solenoid due to the field ?  
• b) If the solenoid is free to turn about the vertical direction, specify its orientations of stable and unstable equilibriums. What is the amount of work needed to displace the solenoid from its stable orientation to its unstable orientation ?

$\underline{\underline{\maltese\:\:\textbf{\textsf{Given}}}}$

• Number of Turns in , N = 1000 turns
• Area of cross section, A = $\sf{2 \times 10^{-4} }$ m²
• Current in closely wound solenoid, I = 2.0 A
• Horizontal axis, $\theta$ = 30°
• Magnetic field, B = 0.16 T

$\underline{\underline{\maltese\:\:\textbf{\textsf{To Find}}}}$

• a) Torque experienced by the solenoid due to the field  
• b) Amount of work needed to displace the solenoid from its stable orientation to its unstable orientation, If the solenoid is free to turn about the vertical direction, specify its orientations of stable and unstable equilibriums

$\underline{\underline{\maltese\:\:\textbf{\textsf{Answer}}}}$

• a) Torque experienced by the solenoid due to the field is 0.032 Nm
• b) The amount of work needed to displace the solenoid from its stable orientation to its unstable orientation is 0.128 J

$\underline{\underline{\maltese\:\:\textbf{\textsf{Calculations}}}}$

a) Torque experienced by the solenoid due to the field

The solenoid is placed in an uniform horizontal magnetic field, the torque experienced by the solenoid is $\sf{\tau=m B \sin \theta}$

• Where m is the magnetic dipole moment of the loop

$:\implies\sf{\tau=m B \sin \theta}$

$:\implies\sf{\tau=m\times 0.16 \:T\times \sin (30^\circ)}$

$:\implies\sf{\tau=N(IA)\times 0.16 \:T\times \sin (30^\circ)}$

$:\implies\sf{\tau=N(IA)\times 0.16 \:T\times 1/2}$

$:\implies\sf{\tau=1000\times(2.0 \:A\times 2\times 10^{-4}\:m^2)\times 0.16 \:T\times 1/2}$

$:\implies\sf{\tau=1000\times(4 \:A\times 10^{-4}\:m^2)\times 0.16 \:T\times 1/2}$

$:\implies\sf{\tau=500\times(4 \:A\times 10^{-4}\:m^2)\times 0.16 \:T\times 1}$

$:\implies\sf{\tau=80\times 4 \:A\times 10^{-4}\:m^2\times T}$

$:\implies\sf{\tau=320 \:A\times 10^{-4}\:m^2\times T}$

$:\implies\sf{\tau=0.032 \:Nm}\implies\boxed{\bf{\tau=0.032 \:Nm}}$

Hence from the Calculations, We can say that :

• The torque experienced by the solenoid due to the field is 0.032 Nm

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b) The amount of work needed to displace the solenoid from its stable orientation to its unstable orientation

$\longrightarrow$ The work done in the magnet is stored in the form of potential energy U

Stable equilibrium : For stable equilibrium, the magnetic moment is parallel to the magnetic moment B and $\sf{\theta=0}$ and $\sf{\tau=0}$ . Therefore, $\sf{U=-m B \cos \theta}$

$:\implies\sf{U_i=-m B \cos \theta}$

$:\implies\sf{U_i=-(0.4\:A\times m^2)\times B\times \cos(0)}$

$:\implies\sf{U_i=-0.4\:A\times m^2\times0.16 \:T\times 1}$

$:\implies\sf{U_i=-0.064\:A\times m^2\times T}$

$:\implies\sf{U_i=-0.064\:J}$

Unstable equilibrium : For Unstable equilibrium, the magnetic moment is anti parallel to the magnetic moment B and $\sf{\theta=180^\circ}$ and $\sf{\tau=0}$

$:\implies\sf{U_f=-m B \cos \theta}$

$:\implies\sf{U_f=-(0.4\:A\times m^2)\times B\times \cos(180^\circ)}$

$:\implies\sf{U_f=-0.4\:A\times m^2\times0.16 \:T\times -1}$

$:\implies\sf{U_f=-0.064\:A\times m^2\times T\times -1}$

$:\implies\sf{U_f=-0.064\:J\times -1}$

$:\implies\sf{U_f=0.064\:J}$

The amount of work done, (The amount of work needed to displace the solenoid from its stable orientation to its unstable orientation) W = $\sf{U_{f}-U_{i}}$

$:\implies\sf{W=U_{f}-U_{i}}$

$:\implies\sf{W=0.064\:J-(-0.064\:J)}$

$:\implies\sf{W=0.064\:J+0.064\:J}$

$\boxed{:\implies\bf{W=0.128\:J}}$

Hence from the Calculations, We can say that :

• The amount of work needed to displace the solenoid from its stable orientation to its unstable orientation is 0.128 J

$\underline{\underline{\maltese\:\:\textbf{\textsf{Request}}}}$

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Answer 2

Given:-

⇒Number of turns(N) = 1000

⇒Current ( I ) = 2A

⇒Area of Cross section = 2×10⁻⁴m²

⇒Magnetic field ( B ) = 0.16T

First of all we find

⇒Magnetic Movement (M) = N(IA)

⇒M = 1000( 2 × 2×10⁻⁴ )

⇒M = 10³ × 2×10⁻⁴ × 2

⇒M = 4 × 10⁻¹ Am²

a) Now we have to find Torque(τ)

Torque(τ) = MBSinФ

⇒τ = MBsin30°

⇒τ = 4×10⁻¹ × 0.16 ×1/2

⇒τ = 0.4 × 0.16 × 1/2

⇒τ = 0.4×0.08

⇒τ = 0.032Nm

b) we have to find amount of work done in stable and unstable orientation

For Stable

τ = 0 and Magnet make 0° to Magnetic field its means parallel

⇒U₁= -MBCosФ

Where Ф = 0°

⇒U₁ = - 0.4×0.16 = 0.064J

For Unstable

τ = 0 and Magnet make 180° to Magnetic field its means anti-parallel

⇒U₂ = -MBCosФ

Where Ф = 180°

⇒U₂ = -(-1)(0.4)(0.160 = 0.064J

The Amount of work ΔW = U₂ - U₁

⇒W = 0.064 - (- 0.064)

⇒W = 0.128J