(A) The magnetic field $B_A$ produced by the infinitely long wire $A$ at a distance $r$ is given by $B_A = \frac{\mu_0 I_A}{2 \pi r}$.Here,$I_A = 10\,

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(A) The magnetic field $B_A$ produced by the infinitely long wire $A$ at a distance $r$ is given by $B_A = \frac{\mu_0 I_A}{2 \pi r}$.Here,$I_A = 10\, A$ and $r = 10\, cm = 0.1\, m$.So,$B_A = \frac{4\pi \times 10^{-7} \times 10}{2\pi \times 0.1} = 2 \times 10^{-5}\, T$.The force $F$ on a current-carrying wire $B$ of length $L$ in a magnetic field $B_A$ is given by $F = I_B L B_A \sin(\theta)$.Since the wires are parallel,the angle $\theta = 90^\circ$ (between the current direction and the magnetic field lines),so $\sin(90^\circ) = 1$.Given $I_B = 2\, A$ and $L = 2\, m$,we have $F = 2 \times 2 \times (2 \times 10^{-5}) = 8 \times 10^{-5}\, N$.

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

Medium

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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A
$8 \times 10^{-5}\, N$
B
$5 \times 10^{-5}\, N$
C
$8\pi \times 10^{-7}\, N$
D
$4\pi \times 10^{-7}\, N$

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Moving Charges and Magnetism

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Force on a Current Carrying Conductor

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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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Two parallel,long wires are kept $0.20 \, m$ apart in a vacuum,each carrying a current of $x \, A$ in the same direction. If the force of attraction per meter of each wire is $2 \times 10^{-6} \, N$,then the value of $x$ is approximately:

In the figure shown,a current $I_1$ is established in the long straight wire $AB$. Another wire $CD$ carrying current $I_2$ is placed in the plane of the paper. The line joining the ends of this wire is perpendicular to the wire $AB$. The force on the wire $CD$ is:

Two wires $A$ and $B$ are carrying currents $I_1$ and $I_2$ as shown in the figure. The separation between them is $d$. $A$ third wire $C$ carrying a current $I$ is to be kept parallel to them at a distance $x$ from $A$ such that the net force acting on it is zero. The possible values of $x$ are

$A$ straight wire of length $50 \text{ cm}$ carrying a current of $2.5 \text{ A}$ is suspended in mid-air by a uniform magnetic field of $0.5 \text{ T}$. Calculate the mass of the wire. (Take $g = 10 \text{ m s}^{-2}$) (in $\text{ g}$)

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