A circular coil having 200 turns, $2.5 \times 10^{-4} \text{ m}^2$ area and carrying 100 $\mu$A current is placed in a uniform magnetic field of 1 T. Initially the magnetic dipole moment ($\vec{M}$) was directed along $\vec{B}$. Amount of work, required to rotate the coil through $90^\circ$ from its initial orientation such that $\vec{M}$ becomes perpendicular to $\vec{B}$, is ________ $\mu$J.
Correct Answer: 5
Approach Solution - 1
We know:
\[ W_{\text{ext}} = \Delta U + \Delta KE \quad (\text{P.E.} = -\vec{M} \cdot \vec{B}) \]
\[ = -MB \cos 90^\circ + MB \cos 0^\circ \]
\[ W_{\text{ext}} = MB \]
\[ = NIAB \]
\[ = 200 \times 100 \times 10^{-6} \times 2.5 \times 10^{-4} \times 1 = 5 \, \mu\text{J} \]
Approach Solution -2
The problem asks for the amount of work required to rotate a current-carrying circular coil in a uniform magnetic field. The coil is initially oriented such that its magnetic dipole moment is parallel to the magnetic field, and it is rotated by 90 degrees to a position where the magnetic dipole moment is perpendicular to the field.
Concept Used:
The work done in rotating a magnetic dipole in a uniform magnetic field is equal to the change in its potential energy. The potential energy (\(U\)) of a magnetic dipole with moment \(\vec{M}\) in a magnetic field \(\vec{B}\) is given by:
\[ U = -\vec{M} \cdot \vec{B} = -MB\cos\theta \]where \(\theta\) is the angle between \(\vec{M}\) and \(\vec{B}\).
The work done (\(W\)) to rotate the dipole from an initial angle \(\theta_1\) to a final angle \(\theta_2\) is:
\[ W = \Delta U = U_{\text{final}} - U_{\text{initial}} = (-MB\cos\theta_2) - (-MB\cos\theta_1) = MB(\cos\theta_1 - \cos\theta_2) \]The magnetic dipole moment (\(M\)) of a coil with \(N\) turns, area \(A\), and carrying current \(I\) is given by:
\[ M = NIA \]Step-by-Step Solution:
Step 1: List the given physical quantities.
Number of turns, \(N = 200\)
Area of the coil, \(A = 2.5 \times 10^{-4} \text{ m}^2\)
Current in the coil, \(I = 100 \text{ }\mu\text{A} = 100 \times 10^{-6} \text{ A} = 10^{-4} \text{ A}\)
Magnetic field strength, \(B = 1 \text{ T}\)
Step 2: Determine the initial and final angles between \(\vec{M}\) and \(\vec{B}\).
Initially, the magnetic dipole moment \(\vec{M}\) is directed along \(\vec{B}\). Therefore, the initial angle is \(\theta_1 = 0^\circ\).
The coil is rotated so that \(\vec{M}\) becomes perpendicular to \(\vec{B}\). Therefore, the final angle is \(\theta_2 = 90^\circ\).
Step 3: Calculate the magnitude of the magnetic dipole moment (\(M\)).
\[ M = NIA \] \[ M = (200) \times (10^{-4} \text{ A}) \times (2.5 \times 10^{-4} \text{ m}^2) \] \[ M = 500 \times 10^{-8} \text{ A}\cdot\text{m}^2 = 5 \times 10^{-6} \text{ A}\cdot\text{m}^2 \]Step 4: Calculate the work done using the formula \(W = MB(\cos\theta_1 - \cos\theta_2)\).
Substitute the values of \(M, B, \theta_1,\) and \(\theta_2\):
\[ W = (5 \times 10^{-6} \text{ A}\cdot\text{m}^2) \times (1 \text{ T}) \times (\cos(0^\circ) - \cos(90^\circ)) \] \[ W = (5 \times 10^{-6}) \times (1 - 0) \] \[ W = 5 \times 10^{-6} \text{ J} \]Final Computation & Result:
The problem asks for the answer in microjoules (\(\mu\)J). Since \(1 \text{ }\mu\text{J} = 10^{-6} \text{ J}\), we can convert the work done from Joules to microjoules.
\[ W = 5 \times 10^{-6} \text{ J} = 5 \text{ }\mu\text{J} \]The amount of work required to rotate the coil is 5 \(\mu\)J.
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