(B) The resistance of the wire is given by $R = \frac{\rho \ell}{A}$.Given $R = 1 \, \Omega$, $\rho = 2 \times 10^{-6} \, \Omega m$, and $A = 10 \, mm

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(B) The resistance of the wire is given by $R = \frac{\rho \ell}{A}$.Given $R = 1 \, \Omega$, $\rho = 2 \times 10^{-6} \, \Omega m$, and $A = 10 \, mm^2 = 10 \times 10^{-6} \, m^2 = 10^{-5} \, m^2$.Substituting these values: $1 = \frac{2 \times 10^{-6} \times \ell}{10^{-5}} \Rightarrow 1 = 0.2 \times \ell \Rightarrow \ell = 5 \, m$.For the wire to be suspended in mid-air, the magnetic force must balance the gravitational force: $F_m = F_g$.$Bi\ell = mg$.Given $i = 2 \, A$, $m = 500 \, g = 0.5 \, kg$, $g = 10 \, m/s^2$, and $\ell = 5 \, m$.$B \times 2 \times 5 = 0.5 \times 10$.$10B = 5$.$B = 0.5 \, T = 5 \times 10^{-1} \, T$.Thus, the magnitude is $5$.

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

Difficult

$A$ $2 \, A$ current carrying straight metal wire of resistance $1 \, \Omega$, resistivity $2 \times 10^{-6} \, \Omega m$, area of cross-section $10 \, mm^2$ and mass $500 \, g$ is suspended horizontally in mid-air by applying a uniform magnetic field $\vec{B}$. The magnitude of $B$ is . . . . . . . $\times 10^{-1} \, T$ (given, $g=10 \, m/s^2$).

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A
$4$
B
$5$
C
$6$
D
$7$

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Moving Charges and Magnetism

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Force on a Current Carrying Conductor

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$A$ current $i = 2 \, A$ flows through a wire frame as shown in the figure. The frame is a combination of two equilateral triangles $ACD$ and $CDE$ with side length $1 \, m$. It is placed in a uniform magnetic field $B = 4 \, T$ acting perpendicular to the plane of the frame. The magnitude of the net magnetic force acting on the frame is.......$N$.

Medium

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Currents of $10\, A$ and $2\, A$ are passed through two parallel thin wires $A$ and $B$ respectively in opposite directions. Wire $A$ is infinitely long and the length of wire $B$ is $2\, m$. The force acting on the conductor $B$,which is situated at a distance of $10\, cm$ from $A$,will be:

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