Subjective Type

Calculate the magnetic moment of a thin wire with a current $$I = 0.8\ A$$, wound tightly on half a tore (Fig.). The diameter of the cross-section of the tore is equal to $$d = 5.0\ cm$$, the number of turns is $$N = 500$$.

Solution

Take an element of length $$rd\theta$$ containing $$\displaystyle\frac{N}{\pi r}\cdot rd\theta$$ turns. Its magnetic moment is

$$\displaystyle\frac{N}{\pi}d\theta\cdot\frac{\pi}{4}d^2I$$

normal to the plane of cross section. We resolve it along $$OA$$ and $$OB$$. The moment along $$OA$$ integrates to

$$\displaystyle\int_0^{\displaystyle\pi}{\frac{N}{4}d^2Id\theta\cos{\theta}}=0$$

while that along $$OB$$ gives

$$\displaystyle p_m=\int_0^{\displaystyle\pi}{\frac{Nd^2I}{4}}\sin{\theta}d\theta=\frac{1}{2}Nd^2I$$


SIMILAR QUESTIONS

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A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 15.0 cm has 500 turns and carries a current of 0.800A.What is the magnetic field inside the toroid at (a) the inner radius and (b) the outer radius?

Magnetism

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