Question 13

A massless square loop, of wire resistance $$10$$ $$\Omega$$, supporting a mass of $$1$$ g, hangs vertically with one of its sides in a uniform magnetic field of $$10^3$$ G, directed outwards in the shaded region. A dc voltage $$V$$ is applied to the loop. For what value of $$V$$, the magnetic force will exactly balance the weight of the supporting mass of $$1$$ g?
(If sides of the loop $$= 10$$ cm, $$g = 10$$ m s$$^{-2}$$)

$$F_{\text{gravity}} = mg = (1 \times 10^{-3}\text{ kg}) \times 10\text{ m/s}^2 = 10^{-2}\text{ N}$$

$$F_{\text{magnetic}} = \left(\frac{V}{R}\right) \cdot l \cdot B$$

$$\frac{V \cdot l \cdot B}{R} = mg$$

$$V = \frac{mg \cdot R}{l \cdot B}$$

$$V = \frac{10^{-2} \times 10}{0.1 \times 0.1}$$

$$V = \frac{10^{-1}}{10^{-2}} = 10\text{ V}$$

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