JEE MAIN - Physics (2023 - 30th January Evening Shift - No. 16)
A current carrying rectangular loop PQRS is made of uniform wire. The length $P R=Q S=5 \mathrm{~cm}$ and $P Q=R S=100 \mathrm{~cm}$. If 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 $\left(f_{P Q}^I: f_{P Q}^{2 t}\right)$ is:
_30th_January_Evening_Shift_en_16_1.png)
1 : 4
1 : 3
1 : 2
1 : 5
Explanation
Force between two current carrying wire
$$ = {{{\mu _0}{I_1}{I_2}} \over {2\pi d}} \times L$$
Here, $${I_1}$$ & $${I_2}$$ are equal
$$F = {{{\mu _0}{I^2}} \over {2\pi d}} \times L$$
$$F \propto {I^2}$$
$${{{F_I}} \over {{F_{2I}}}} = {{{I^2}} \over {4{I^2}}} = {1 \over 4}$$
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