(C) The magnetic force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $d$ is given by $f = \frac{\

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(C) The magnetic force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $d$ is given by $f = \frac{\mu_0 I_1 I_2}{2 \pi d}$.For length $L$,the force is $F = \frac{\mu_0 I^2 L}{2 \pi d}$.In the new condition,the distance $d' = \frac{d}{2}$ and the currents $I' = 2I$.The new force $F_2$ is given by $F_2 = \frac{\mu_0 (2I)(2I) L}{2 \pi (d/2)}$.Simplifying this,we get $F_2 = \frac{\mu_0 (4I^2) L}{2 \pi (d/2)} = 8 \times \left( \frac{\mu_0 I^2 L}{2 \pi d} \right)$.Therefore,$F_2 = 8F$.

Class 12 Physics Moving Charges and Magnetism Force on a Current Carrying Conductor

Medium

Two long straight wires $A$ and $B$ carrying equal current $I$ were kept parallel to each other at distance $d$ apart. The magnitude of the magnetic force experienced by length $L$ of wire $A$ is $F$. If the distance between the wires is halved and the currents are doubled,the force $F_2$ on length $L$ of wire $A$ will be:

MHT CET 2025 All MHT CET

A
$2 F$
B
$F$
C
$8 F$
D
$4 F$

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Moving Charges and Magnetism

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Force on a Current Carrying Conductor

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$A$ rigid square loop of side $a$ carrying current $I_2$ is lying on a horizontal surface near a long wire carrying current $I_1$ in the same plane as shown in the figure. The net force on the loop due to the wire will be:

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$A$ square loop $ABCD$ of side $L$ carrying a current $I_1$ is placed at a distance $(L/3)$ from a straight conductor $XY$ carrying current $I_2$. The loop and the conductor are coplanar. The net force on the loop will be ($\mu_0$ = magnetic permeability).

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$A$ semi-circular loop of radius $30 \,cm$ wire carries current $6 \,A$. $A$ uniform magnetic field $0.5 \,T$ is present perpendicular to the plane of the loop. What is the magnitude of force exerted on the wire (in $\,N$)?

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An isosceles triangular current-carrying loop is placed in a uniform magnetic field $\overrightarrow{B_{o}}$ directed perpendicular to the plane of the loop,as shown in the figure. The net magnetic force on the loop is:

$ABCD$ is a rectangular loop made of uniform wire. If $AD = BC = 2 \text{ cm}$,what is the magnetic force per unit length acting on wire $DC$ due to wire $AB$ if the ammeter reads $20 \text{ A}$? (The lengths of $AB$ and $DC$ are large in comparison with the other two sides).

$A$ rigid square loop of side $a$ carrying current $I_2$ is lying on a horizontal surface near a long wire carrying current $I_1$ in the same plane as shown in the figure. The net force on the loop due to the wire will be:

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