(C) The force due to gravity acting down the incline is $F_g = mg \sin 30^{\circ}$.The magnetic force acting on the rod is $F_m = IlB$,which acts hori

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(C) The force due to gravity acting down the incline is $F_g = mg \sin 30^{\circ}$.The magnetic force acting on the rod is $F_m = IlB$,which acts horizontally. The component of this magnetic force acting up the incline is $F_{m, \text{up}} = IlB \cos 30^{\circ}$.For the rod to remain stationary,the forces along the incline must balance:$mg \sin 30^{\circ} = IlB \cos 30^{\circ}$Rearranging for current $I$:$I = \frac{mg}{lB} \tan 30^{\circ}$Given mass per unit length $\frac{m}{l} = 0.5\; kg\; m^{-1}$,$B = 0.25\; T$,and $g = 9.8\; m/s^2$:$I = \frac{0.5 \times 9.8}{0.25} \times \tan 30^{\circ}$$I = 2 \times 9.8 \times \frac{1}{\sqrt{3}}$$I = \frac{19.6}{1.732} \approx 11.32\; A$

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

Medium

$A$ metallic rod of mass per unit length $0.5\; kg\; m^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^{\circ}$ with the horizontal. The rod is prevented from sliding down by passing a current through it,while a magnetic field of induction $0.25\; T$ acts on it in the vertical direction. The current flowing in the rod to keep it stationary is: (in $; A$)

NEET 2018 All NEET

A
$7.14$
B
$5.98$
C
$11.32$
D
$14.76$

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Class 12

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Moving Charges and Magnetism

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Force on a Current Carrying Conductor

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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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$A$ thin flexible wire of length $L$ is connected to two adjacent fixed points and carries a current $I$ in the clockwise direction,as shown in the figure. When the system is put in a uniform magnetic field of strength $B$ going into the plane of the paper,the wire takes the shape of a circle. The tension in the wire is

AP EAMCET 2021

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$A$ straight wire of mass $0.2 \,kg$ and length $1.5 \,m$ carries a current of $2 \,A$, as shown in the figure. It is suspended in mid-air by a uniform magnetic field $B$ pointing into the plane of the paper. Calculate the magnitude of the magnetic field. (Ignore Earth's magnetic field and assume $g = 10 \,m/s^2$) (in $\,T$)

MediumTS EAMCET 2020

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The figure shows a cross-section of an infinite metal sheet carrying an electric current along its surface. The current per unit length is $J$. $A$ current-carrying square loop is placed nearby the metal sheet such that the plane of the square is perpendicular to the plane of the sheet. Then:

Medium

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What is the magnitude of magnetic force per unit length (in $N \;m^{-1}$) on a wire carrying a current of $8 \;A$ and making an angle of $30^{\circ}$ with the direction of a uniform magnetic field of $0.15 \;T$?

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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).

The figure shows a cross-section of an infinite metal sheet carrying an electric current along its surface. The current per unit length is $J$. $A$ current-carrying square loop is placed nearby the metal sheet such that the plane of the square is perpendicular to the plane of the sheet. Then:

What is the magnitude of magnetic force per unit length (in $N \;m^{-1}$) on a wire carrying a current of $8 \;A$ and making an angle of $30^{\circ}$ with the direction of a uniform magnetic field of $0.15 \;T$?

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