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6. ⇒ (MHT CET 2023 9th May Morning Shift )

A long straight wire carrying a current of $25 A$ rests on the table. Another wire PQ of length $1 m$ and mass $2.5 g$ carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will wire PQ rise? ( $μ_{0} = 4 π \times 10^{- 7}$ SI unit)

MHT CET 2023 9th May Morning Shift Physics - Moving Charges and Magnetism Question 20 English

A. 3 mm

B. 4 mm

C. 5 mm

D. 8 mm

Correct Option is (C)

Given $I_{1} = I_{2} = 25 A, l = 1 m$ ,

$\begin{aligned} B = \frac{μ_{0} I}{2 π h} \\ F = BI l \sin θ = BI l \end{aligned}$

Force applied on $PQ =$ Weight of the smaller current carrying wire

$\begin{array}{ll} i.e, mg = \frac{μ_{0} I^{2} l}{2 π h} \\ ∴ h = \frac{4 π \times 10^{- 7} \times 250 \times 25 \times 1}{2 π \times 2.5 \times 10^{- 3} \times 9.8} = 5 mm \end{array}$

7. ⇒ (MHT CET 2021 21th September Evening Shift )

A, B and C are three parallel conductors of equal lengths carrying currents $I, I$ and $2 I$ respectively. Distance between A and B is ' $x$ ' and that between B and C is also ' $x$ '. $F_{1}$ is the force exerted by conductor $B$ on $A$ . $F_{2}$ is the force exerted by conductor $C$ on $A$ . Current $I$ in $A$ and $I$ in $B$ are in same direction and current $2 I$ in $C$ is in opposite direction. Then

A. $F_{1} = F_{2}$

B. $F_{2} = 2 R_{1}$

C. $F_{1} = 2 R_{2}$

D. $F_{1} = - F_{2}$

Correct Option is (D)

MHT CET 2021 21th September Evening Shift Physics - Moving Charges and Magnetism Question 24 English Explanation

Currents in $A$ and $B$ are in the same direction. Hence force $F_{1}$ exerted by $B$ on $A$ will attractive (towards B). Current in $A$ and $C$ are in opposite directions. Hence force $F_{2}$ , exentos by $C$ on $A$ will be repulsive (away from $C$ ). Thus $F_{1}$ and $F_{2}$ are opposite in direction.

$\begin{aligned} F_{1} = \frac{μ_{0}}{2 π} \cdot \frac{I_{1} I_{2}}{x} \cdot L = \frac{μ_{0}}{2 π} \cdot \frac{I^{2}}{x} \cdot L \\ F_{2} = \frac{μ_{0}}{2 π} \cdot \frac{2 I^{2}}{2 x} \cdot L = \frac{μ_{0}}{2 π} \cdot \frac{I^{2}}{x} \cdot L \end{aligned}$

$∴ F_{1}$ and $F_{2}$ have some magnitude.

$∴ F_{1} = - F_{2}$

⇒Magnetic Field due to Electric Current