(B) The forces acting on the rod are the gravitational force $mg$ (downwards) and the magnetic force $F_m = ilB$ (upwards,by Fleming's Left-Hand Rule)

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(B) The forces acting on the rod are the gravitational force $mg$ (downwards) and the magnetic force $F_m = ilB$ (upwards,by Fleming's Left-Hand Rule).The net force on the rod is $F_{net} = ilB - mg$ (upwards).Since the rod is in equilibrium,the two springs must provide a restoring force equal to the net force. Let $x$ be the extension in each spring. The total restoring force provided by the two springs is $2Kx$.Equating the forces: $2Kx = ilB - mg$.Solving for the extension $x$: $x = \frac{ilB - mg}{2K}$.The length of each spring in the equilibrium state is $l = l_0 + x$.Substituting the value of $x$: $l = l_0 + \frac{ilB - mg}{2K}$.

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

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$A$ horizontal metallic rod of mass $m$ and length $l$ is supported by two vertical identical springs of spring constant $K$ each and natural length $l_0$. $A$ current $i$ is flowing in the rod in the direction shown. If the rod is in equilibrium,then the length of each spring in this state is:

A
$l_0 + \frac{ilB - mg}{K}$
B
$l_0 + \frac{ilB - mg}{2K}$
C
$l_0 + \frac{mg - ilB}{2K}$
D
$l_0 + \frac{mg - ilB}{K}$

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Moving Charges and Magnetism

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Force on a Current Carrying Conductor

The magnetic field existing in a region is given by $\vec B = B_0 \left( 5 + \frac{x}{l} \right) \hat k$. $A$ square loop of edge $l$ and carrying a current $i$ is placed with its edges parallel to $x-y$ axes. Find the magnitude of the net magnetic force experienced by the loop.

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$A$ wire of length $L$ carries current $I$ along the $x$-axis. $A$ magnetic field $\vec{B} = B_0(\hat{i} - \hat{j} - \hat{k}) \text{ T}$ acts on the wire. The magnitude of the magnetic force acting on the wire is:

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$A$ loop of flexible conducting wire lies in a magnetic field of $2.0 \,T$ with its plane perpendicular to the field. The length of the wire is $1 \,m$. When a current of $1.1 \,A$ is passed through the loop, it opens into a circle. The tension developed in the wire is (in $\,N$)

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$A$ square loop of edge length $2 \ m$ carrying current of $2 \ A$ is placed with its edges parallel to the $x-y$ axis. $A$ magnetic field is passing through the $x-y$ plane and expressed as $\vec{B}=B_0(1+4x) \hat{k}$,where $B_0=5 \ T$. The net magnetic force experienced by the loop is . . . . . . $N$.

DifficultJEE MAIN 2024

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The magnetic field existing in a region is given by $\vec{B} = 0.2(1 + 2x) \hat{k} \text{ T}$. $A$ square loop of edge $50 \text{ cm}$ carrying $0.5 \text{ A}$ current is placed in the $x-y$ plane with its edges parallel to the $x-y$ axes,as shown in the figure. The magnitude of the net magnetic force experienced by the loop is . . . . . . $\text{mN}$.

DifficultJEE MAIN 2024

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$A$ horizontal metallic rod of mass $m$ and length $l$ is supported by two vertical identical springs of spring constant $K$ each and natural length $l_0$. $A$ current $i$ is flowing in the rod in the direction shown. If the rod is in equilibrium,then the length of each spring in this state is:

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The magnetic field existing in a region is given by $\vec B = B_0 \left( 5 + \frac{x}{l} \right) \hat k$. $A$ square loop of edge $l$ and carrying a current $i$ is placed with its edges parallel to $x-y$ axes. Find the magnitude of the net magnetic force experienced by the loop.

$A$ wire of length $L$ carries current $I$ along the $x$-axis. $A$ magnetic field $\vec{B} = B_0(\hat{i} - \hat{j} - \hat{k}) \text{ T}$ acts on the wire. The magnitude of the magnetic force acting on the wire is:

$A$ loop of flexible conducting wire lies in a magnetic field of $2.0 \,T$ with its plane perpendicular to the field. The length of the wire is $1 \,m$. When a current of $1.1 \,A$ is passed through the loop, it opens into a circle. The tension developed in the wire is (in $\,N$)

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