Force on a Current Carrying Conductor in Magnetic Field

A metal ring of radius r = 0.5 m with its plane normal to a uniform magnetic field B of induction 0.2 T carries a current I = 100 A. The tension in newtons developed in the ring is :

A semi circular current carrying wire having radius $R$ is placed in $\mathrm{x}-\mathrm{y}$ plane with its centre at origin $\mathrm{{\rm O}}$. There is non-uniform magnetic field $\overrightarrow{B}=\left(\frac{{\mathrm{B}}_{\mathrm{o}}\mathrm{x}}{\mathrm{2}\mathrm{R}}\right)\widehat{\mathrm{k}}$ $\left(\text{here}{B}_{\mathrm{o}} \text{is} +\text{ve constant}\right)$  is existing in the region. The magnetic force acting on semi-circular wire will be along:

A conducting wire bent in the form of a parabola ${\text{y}}^2=2\text{x}$  carries a current i = 2 A as shown in figure. This wire is placed in a uniform magnetic field $\overrightarrow{\text{B}}=-\text{4}\widehat{\text{k}}$ Tesla. The magnetic force on the wire is (in newton)

A straight rod of mass m and length $L$ is suspended from the identical spring as shown in the figure. The spring stretched by a distance of ${\mathit{\text{x}}}_0$ due to the weight of the wire. The circuit has total resistance $R \mathrm{\Omega }$. When the magnetic field perpendicular to the plane of the paper is switched on, springs are observed to extend further by the same distance. The magnetic field strength is

A long thin-walled pipe of radius $R$ carries a current $i$ along its length. The current density is uniform over the circumference of the pipe. The magnetic field at the centre of the pipe due to the quarter portion of the pipe shown, is

An infinite wire placed along $z$-axis, has current $I_1$ in positive $z$-direction. A conducting rod placed in $xy$ plane parallel to y-axis has current $I_2$ in positive y-direction. The ends of the rod subtend angles of $+30°$ and $-60°$ at the origin with positive $x$-direction. The rod is at a distance $a$ from the origin. Find net force on the rod.

A $U$ shaped wire of mass $m$ and length $l$ is immersed with its two ends in mercury (see figure). The wire is in a homogeneous field of magnetic induction $B$. If a charge, that is, a current pulse $q=\int i\mathrm{d}t,$sent through the wire, the wire will jump up. Calculate, from the height $h$ that the wire reaches, the size of the charge or current pulse, assuming that the time of the current pulse is very small in comparison with the time of flight. Make use of the fact that impulse of force equals $\int F\mathrm{d}t,$ which equals $mv$. Evaluate $q$ for following details:
$B=0.1 \mathrm{Wb} {\mathrm{m}}^{-2}\mathrm{,} m=10 \mathrm{g}\mathrm{,}l=20 \mathrm{cm}$
$\text{\&} \mathit{h}=3 \mathrm{m}.\left[g=\mathrm{10} \mathrm{m} {\mathrm{s}}^{-2}\right]$

A rectangular loop of wire is oriented with the left corner at the origin, one edge along X-axis and the other edge along Y-axis as shown figure. A magnetic field is into the page and has a magnitude that is given by b=$\alpha \text{y}\text{where}\alpha \text{is }$constant. Find the total magnetic force on the loop if it carries current i.

An arc of a circular loop of radius $R$ is kept in the horizontal plane and a
constant magnetic field $B$ is applied in the vertical direction as shown in
the figure. If the arc carries current $I$ then find the force on the arc.

Electric charge $q$ is uniformly distributed over a rod of length $l$. The rod is placed parallel to a long wire carrying a current $i$. The separation between the rod and the wire is $a$. Find the force needed to move the rod along its length with a uniform velocity $v$.

Figure shows a rectangular current-carrying loop placed $2 cm$ away from a long, straight, current-carrying conductor. What is the magnitude of the net force acting on the loop in ${10}^{-4}N$.

The horizontal component of the earth's magnetic field at a certain place is $3.0\times {10}^{-5} \mathrm{T}$ and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of $2 \mathrm{A}$. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is east to west?

A current carrying wire is placed parallel to the lines of force in a magnetic field.

The horizontal component of Earth’s magnetic field at a certain place is $3.0\times {10}^{-5}\mathrm{ }\mathrm{T}$. It is directed from the geographic south to the geographic north. The force per unit length on a very long straight conductor carrying a steady current of $1.2\mathrm{ }\mathrm{A}$ in east-west direction is:

A current of $10 \mathrm{A}$ is flowing in a wire of length $1.5 \mathrm{m}$. A force of $15 \mathrm{N}$ acts on it when it is placed in a uniform magnetic field of $2 \mathrm{T}$. The angle between the magnetic field and the direction of the current is:

Consider two infinite long parallel current-carrying sheets. Current per unit width in both sheets is $\sqrt{\frac{{10}^7}{\pi }}\mathrm{A}/\mathrm{m}$ and direction of current in both sheets is same then calculate force (in $\mathrm{N}$) per unit area on each sheet.

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

Figure shows an equilateral triangle $ABC$ of side $l$ carrying currents placed in uniform magnetic field $B$. The magnitude of magnetic force on triangle is,

A conduction rod of $1 \mathrm{m}$ length and $1 \mathrm{kg}$ mass is suspended by two vertical wires through its ends. An external magnetic field of $2 \mathrm{T}$ is applied normal to the rod. Now the current to be passed through the rod so as to make the tension in the wires zero is

$\left[\mathrm{T}\mathrm{a}\mathrm{k}\mathrm{e}\mathrm{ }\mathrm{g}=10\mathrm{ }{\mathrm{m}\mathrm{s}}^{-2}\right]$