Question

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

Answer 1

Correct Answer: 100

The problem asks for the work done in rotating a current-carrying coil in a uniform magnetic field from an initial orientation to a final one. We are given the coil's properties, the current it carries, the strength of the magnetic field, and the angle of rotation.

Concept Used:

The work done in rotating a magnetic dipole (such as a current-carrying coil) in a uniform magnetic field is equal to the change in its potential energy. The potential energy \(U\) of a magnetic dipole in a magnetic field is given by the dot product of the magnetic moment and the magnetic field:

\[ U = -\vec{\mu} \cdot \vec{B} = -\mu B \cos\theta \]

where \(\mu\) is the magnitude of the magnetic dipole moment, \(B\) is the magnitude of the magnetic field, and \(\theta\) is the angle between the magnetic moment vector \(\vec{\mu}\) and the magnetic field vector \(\vec{B}\).

The magnetic moment of a coil with \(N\) turns, area \(A\), and carrying current \(I\) is:

\[ \mu = NIA \]

The work done \(W\) in changing the orientation of the coil from an initial angle \(\theta_i\) to a final angle \(\theta_f\) is:

\[ W = \Delta U = U_f - U_i = (-\mu B \cos\theta_f) - (-\mu B \cos\theta_i) = \mu B (\cos\theta_i - \cos\theta_f) \]

Step-by-Step Solution:

Step 1: Determine the initial and final angles (\(\theta_i\) and \(\theta_f\)).

The magnetic moment vector \(\vec{\mu}\) is always perpendicular to the plane of the coil. The initial state is given as "the plane of the coil is perpendicular to the magnetic field". This means the magnetic moment vector \(\vec{\mu}\) is parallel to the magnetic field vector \(\vec{B}\). Therefore, the initial angle is:

\[ \theta_i = 0^\circ \]

The coil is then turned through \(90^\circ\). The final angle is:

\[ \theta_f = \theta_i + 90^\circ = 0^\circ + 90^\circ = 90^\circ \]

Step 2: Calculate the magnitude of the magnetic dipole moment (\(\mu\)).

We are given:

• Number of turns, \(N = 100\)
• Area, \(A = 5 \times 10^{-3} \, \text{m}^2\)
• Current, \(I = 1 \, \text{mA} = 1 \times 10^{-3} \, \text{A}\)

Using the formula \(\mu = NIA\):

\[ \mu = (100) \times (1 \times 10^{-3} \, \text{A}) \times (5 \times 10^{-3} \, \text{m}^2) \] \[ \mu = 500 \times 10^{-6} \, \text{A} \cdot \text{m}^2 = 5 \times 10^{-4} \, \text{A} \cdot \text{m}^2 \]

Step 3: Calculate the work done using the formula.

The work done is \(W = \mu B (\cos\theta_i - \cos\theta_f)\). We have:

• \(\mu = 5 \times 10^{-4} \, \text{A} \cdot \text{m}^2\)
• Magnetic field, \(B = 0.20 \, \text{T}\)
• \(\cos\theta_i = \cos(0^\circ) = 1\)
• \(\cos\theta_f = \cos(90^\circ) = 0\)

Substituting these values:

\[ W = (5 \times 10^{-4}) \times (0.20) \times (1 - 0) \] \[ W = 1 \times 10^{-4} \, \text{J} \]

Step 4: Convert the work done to microjoules (\(\mu\text{J}\)).

Since \(1 \, \mu\text{J} = 10^{-6} \, \text{J}\), we can convert the result:

\[ W = 1 \times 10^{-4} \, \text{J} = (1 \times 10^{-4}) \times 10^6 \, \mu\text{J} = 100 \, \mu\text{J} \]

The work done in turning the coil is 100 \(\mu\text{J}\).

Answer 2

The work done \( W = \Delta U = U_f - U_i \):
\[W = -(\mu B)_f - (-\mu B)_i\]
Initially, the magnetic moment \( \mu \) is perpendicular to the magnetic field, so:
\[W = 0 + (\mu B)\]
Substitute the values:
\[\mu = (100 \times 5 \times 10^{-3} \times 1 \times 10^{-3}) \, \text{A} \cdot \text{m}^2\]
\[W = (1 \times 10^{-4}) \times 0.2 \, \text{J} = 1 \times 10^{-5} \, \text{J} = 100 \, \mu \text{J}\]