Question

a magnet is parallel to a uniform magnetic field.if it is rotated by 60 degrees,the work done is 0.8 J. how much additional work is done in moving it further through 30 degrees from that position?

Answer 1

Answer on Question #62811, Physics / Electromagnetism

a magnet is parallel to a uniform magnetic field. if it is rotated by 60 degrees, the work done is 0.8 J. how much additional work is done in moving it further through 30 degrees from that position?

Find: $\Delta A - ?$

Given:

$$A_1 = 0.8 \, \text{J}$$ $$\alpha_1 = 60{}^\circ$$ $$\alpha_2 = 90{}^\circ$$

Solution:

$$A \sim \sin \alpha \, (1)$$ $$\begin{array}{l} \text{Of} \, (1) \Rightarrow A_1 \sim \sin \alpha_1 \, (2) \\ \quad A_2 \sim \sin \alpha_2 \, (3) \\ \end{array}$$ $$\begin{array}{l} \text{Of} \, (2) \text{ and } (3) \Rightarrow A_2 = \frac{\sin \alpha_2}{\sin \alpha_1} A_1 \, (4) \\ \text{Of} \, (4) \Rightarrow A_2 = 0.9238 \, \text{J} \\ \Delta A = A_2 - A_1 \, (5) \\ \text{Of} \, (5) \Rightarrow \Delta A = 0.1238 \, \text{J} \\ \end{array}$$

Answer:

0.1238 J

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