To solve the problem of finding the work done in rotating the bar magnet, we need to use the formula that relates the work done in rotating a magnetic moment in a magnetic field. The formula for the work done [tex]\( W \)[/tex] is given by:
[tex]\[ W = m \cdot B \cdot (1 - \cos \theta) \][/tex]
where:
- [tex]\( m \)[/tex] is the magnetic moment,
- [tex]\( B \)[/tex] is the magnetic field strength,
- [tex]\( \theta \)[/tex] is the angle of rotation.
We are provided with the following values:
- The magnetic moment, [tex]\( m = 2.0 \times 10^5 \)[/tex] JT[tex]\(^{-1}\)[/tex],
- The magnetic field, [tex]\( B = 14 \times 10^{-5} \)[/tex] T,
- The angle of rotation, [tex]\( \theta = 60^\circ \)[/tex].
First, we need to convert the angle from degrees to radians. The conversion formula is:
[tex]\[ \text{angle in radians} = \text{angle in degrees} \times \left(\frac{\pi}{180}\right) \][/tex]
Substituting [tex]\( 60^\circ \)[/tex]:
[tex]\[ \theta = 60^\circ \times \left(\frac{\pi}{180}\right) = \frac{\pi}{3} \approx 1.047 \text{ radians} \][/tex]
Next, we compute [tex]\( \cos(\theta) \)[/tex]. For [tex]\( \theta = 60^\circ \)[/tex]:
[tex]\[ \cos(60^\circ) = \cos \left(\frac{\pi}{3}\right) = \frac{1}{2} \][/tex]
Now, substitute all the known values into the work done formula:
[tex]\[ W = m \cdot B \cdot (1 - \cos \theta) \][/tex]
[tex]\[ W = 2.0 \times 10^5 \times 14 \times 10^{-5} \times \left(1 - \frac{1}{2}\right) \][/tex]
Simplify the computation step-by-step:
1. Calculate the product of [tex]\( m \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ 2.0 \times 10^5 JT^{-1} \times 14 \times 10^{-5} T = 2.8 \][/tex]
2. Compute [tex]\( 1 - \cos(60^\circ) \)[/tex]:
[tex]\[ 1 - \frac{1}{2} = \frac{1}{2} \][/tex]
3. Finally, calculate the work done:
[tex]\[ W = 2.8 \times \frac{1}{2} = 1.4 \][/tex]
So, the work done in rotating the bar magnet through [tex]\( 60^\circ \)[/tex] is:
[tex]\[ W = 1.4 \][/tex] J
Therefore, the correct answer is:
(4) 1.4 J
