1. ⇒ (MHT CET 2023 12th May Morning Shift )

$A, B$ and $C$ are three parallel conductors of equal lengths and carry currents I, I and 2I respectively as shown in figure. Distance $A B$ and $B C$ is same as ' $d$ '. If ' $F_{1}$ ' is the force exerted by $B$ on $A$ and $F_{2}$ is the force exerted by $C$ on $A$ , then

MHT CET 2023 12th May Morning Shift Physics - Moving Charges and Magnetism Question 9 English

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

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

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

D. $F_{1} = \frac{1}{2} F_{2}$

Correct Option is (B)

Force per unit length exerted by B on A, $F_{1} = \frac{μ_{0} (I) (I)}{2 π d} = \frac{μ_{0} I^{2}}{2 π d}$ (outside the plane of paper)

Force per unit length exerted by $C$ on $A$ ,

$F_{2} = \frac{μ_{0} (I) (2 I)}{2 π (2 d)} = \frac{μ_{0} I^{2}}{2 π d}$ (Inside the plane of paper)

$∴ F_{1} = - F_{2}$

2. ⇒ (MHT CET 2023 12th May Morning Shift )

A single turn current loop in the shape of a right angle triangle with side $5 cm, 12 cm, 13 cm$ is carrying a current of $2 A$ . The loop is in a uniform magnetic field of magnitude $0.75 T$ whose direction is parallel to the current in the $13 cm$ side of the loop. The magnitude of the magnetic force on the $5 cm$ side will be $\frac{x}{130} N$ . The value of ' $x$ ' is

A. 4

B. 9

C. 12

D. 15

Correct Option is (B)

MHT CET 2023 12th May Morning Shift Physics - Moving Charges and Magnetism Question 8 English Explanation

The net magnetic field is acting in the direction of $GF$ as shown in figures.

Resolving $\vec{B}$ into its components, amongst the components, only $\vec{B} \sin θ$ exerts force on side $EF$ of current carrying loop.

$∴ F_{EF} = I \times d (EF) \times B \sin θ$

From figure (a), $\sin θ = \frac{12}{13}$

$\begin{array}{ll} ∴ & F_{EF} = 2 \times 0.05 \times 0.75 \times \frac{12}{13} \\ ∴ & F_{EF} = \frac{9}{130} N \\ ∴ & x = 9 \end{array}$

3. ⇒ (MHT CET 2023 11th May Evening Shift )

Two long conductors separated by a distance ' $d$ ' carry currents $I_{1}$ and $I_{2}$ in the same direction. They exert a force ' $F$ ' on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance between them is also increased to $3 d$ . The new value of force between them is

A. $- 2 F$

B. $\frac{F}{3}$

C. $\frac{- 2 F}{3}$

D. $\frac{- F}{3}$

Correct Option is (C)

The force per unit length of the conductors is given as:

$F = \frac{μ_{0} I_{1} I_{2}}{2 π d}$

When the value and direction of current in the first conductor and the distance between the conductors are changed,

$\begin{aligned} ∴ F_{2} & = \frac{- μ_{0} 2 I_{1} I_{2}}{2 π \times 3 d} \\ ∴ & \frac{F_{2}}{F} = \frac{- 2}{3} \\ F_{2} & = - \frac{2 F}{3} \end{aligned}$

4. ⇒ (MHT CET 2023 10th May Evening Shift )

Two wires $2 mm$ apart supply current to a $100 V, 1 kW$ heater. The force per metre between the wires is ( $μ_{0} = 4 π \times 10^{- 27}$ SI unit)

A. $2 \times 10^{- 2} N$

B. $4 \times 10^{- 3} N$

C. $2 \times 10^{2} N$

D. $10^{- 2} N$

Correct Option is (D)

$\begin{array}{ll} Given: P = 1 W = 1000 W, V = 100 V \\ ∴ & I = 10 A . . . . (∵ P = VI) \\ ∴ & F = \frac{μ_{0} I_{1} I_{2}}{2 π \times a} = \frac{4 π \times 10^{- 7} \times 100}{2 π \times (2 \times 10^{- 3})} = 10^{- 7} \times 10^{5} = 10^{- 2} N \end{array}$

5. ⇒ (MHT CET 2023 9th May Evening Shift )

10 A current is flowing in two straight parallel wires in the same direction. Force of attraction between them is $1 \times 10^{- 3} N$ . If the current is doubled in both the wires the force will be

A. $1 \times 10^{- 3} N$

B. $2 \times 10^{- 3} N$

C. $4 \times 10^{- 3} N$

D. $0.25 \times 10^{- 3} N$

Correct Option is (C)

$F = \frac{μ_{0}}{4 π} \frac{2 I_{1} I_{2}}{r} l = 10^{- 3} N$

When current in both the wires is doubled, then

$F^{'} = \frac{μ_{0}}{4 π} \frac{2 (2 I_{1} \times 2 I_{2})}{r} l = 4 \times 10^{- 3} N$

⇒Magnetic Field due to Electric Current