A current carrying rectangular loop is made of uniform wire. The length cm and cm. If the ammeter current reading changes from to , the ratio of magnetic forces per unit length on the wire due to wire in the two cases respectively is [

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Solution

Q.
A current carrying rectangular loop $P Q R S$ is made of uniform wire. The length $P R = Q S = 5$ cm and $P Q = R S = 100$ cm. If the ammeter current reading changes from $I$ to $2 I$ , the ratio of magnetic forces per unit length on the wire $P Q$ due to wire $R S$ in the two cases respectively $f_{P Q}^{I} : f_{P Q}^{2 I}$ is [2023]

1 1 : 2

2 1 : 4

3 1 : 5

4 1 : 3

Ans.
(2)

$Force between two current-carrying wires = \frac{μ_{0} I_{1} I_{2}}{2 π d} \times L$

$Here I_{1} and I_{2} are equal$

$F = \frac{μ_{0} I^{2}}{2 π d} \times L$

$F \propto I^{2}$

$\frac{F_{I}}{F_{2 I}} = \frac{I^{2}}{4 I^{2}} = \frac{1}{4}$

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