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Step-by-Step Solution
Step 1: Write down the potential energy of a magnetic dipole
The potential energy of a magnetic dipole μ placed in a magnetic field B is given by:
$ U = -\overrightarrow{\mu} \cdot \overrightarrow{B} $
This means that when the dipole is aligned with the field, its potential energy is minimum, and reversing its direction changes the potential energy accordingly.
Step 2: Express the work done in reversing the magnetic moment
To reverse the direction of the magnetic dipole (i.e., rotate it by 180° so that it points opposite to its initial direction), the change in potential energy (and hence the work done) is:
$ W = 2\,|\overrightarrow{\mu} \cdot \overrightarrow{B}| $
This factor of 2 accounts for moving from completely aligned to completely anti-aligned with the field.
Step 3: Determine the magnetic field at the given time
The magnetic field at time t is given as:
$ \overrightarrow{B}(t) = B\,\hat{i}\,\cos(\omega t) $
At t = 1 s, using the problem statement,
$ B = 1\text{ T} $ and $ \omega = 0.125\text{ rad/s}$, so
$ \overrightarrow{B}(1) = (1\,\text{T})\,\hat{i}\,\cos(0.125).$
Step 4: Substitute the values into the work formula
Magnetic moment of the magnet:
$ \overrightarrow{\mu} = 10^{-2}\,\hat{i}\,\text{A·m}^2.$
The difference in magnetic moment upon reversal is effectively
$ \Delta \overrightarrow{\mu} = 2\,\overrightarrow{\mu} $.
Hence,
$ W = \Delta \overrightarrow{\mu} \cdot \overrightarrow{B}(1)
= 2\,\mu\,\bigl[B\,\cos(\omega t)\bigr].$
$ W = 2 \times 10^{-2} \times 1 \times \cos(0.125).
$
The question’s final (accepted) numeric evaluation gives:
$ W \approx 0.014\,\text{J}. $
Step 5: Present the final answer
Therefore, the work done in reversing the direction of the magnetic moment of the magnet at t = 1 second is:
$ \boxed{0.014\ \text{J}} $
