Question 12

A current carrying rectangular loop $$PQRS$$ is made of a uniform wire. The length $$PR = QS = 5$$ cm and $$PQ = RS = 100$$ cm. If the ammeter current reading changes from $$I$$ to $$2I$$, then the ratio of the magnetic forces per unit length on the wire $$PQ$$ due to the wire $$RS$$ in the two cases respectively $$(f^I_{PQ} : f^{2I}_{PQ})$$ is:

The magnetic force per unit length, $$f = \frac{\mu_0 i_1 i_2}{2\pi d}$$

Current in wire $$PQ$$ ($$i_1$$) $$= \frac{I_{total}}{2}$$

Current in wire $$RS$$ ($$i_2$$) $$= \frac{I_{total}}{2}$$

Case 1: Ammeter reading is $$I$$

$$i_1 = \frac{I}{2}$$ and $$i_2 = \frac{I}{2}$$

Force per unit length $$f_1 = \frac{\mu_0 (\frac{I}{2}) (\frac{I}{2})}{2\pi d} = \frac{\mu_0 I^2}{8\pi d}$$

Case 2: Ammeter reading is $$2I$$

$$i_1 = \frac{2I}{2} = I$$ and $$i_2 = \frac{2I}{2} = I$$

Force per unit length $$f_2 = \frac{\mu_0 (I) (I)}{2\pi d} = \frac{\mu_0 I^2}{2\pi d}$$

$$\frac{f_1}{f_2} = \frac{\frac{\mu_0 I^2}{8\pi d}}{\frac{\mu_0 I^2}{2\pi d}} = \frac{2}{8} = \frac{1}{4}$$

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