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Force On A Conductor Carrying Current In A Magnetic Field - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Magnetic force on a current carrying conductor is considered one of the most asked concept.

  • 30 Questions around this concept.

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 A wire carrying current I is tied between points P and Q and is in the shape of a circular arch of radius R due to a uniform magnetic field B (perpendicular to the plane of the paper, shown by xxx) in the vicinity of the wire. If the wire subtends an angle  $2\Theta_0$  at the center of the circle (of which it forms an arch) then the tension in the wire is :

A conductor lies along the Z-axis at    -1.5\leq z< 1.5 \: m and carries a fixed current of 10.0 A in  - \hat{a_{z}} direction (see figure).

for a field \vec{B}=3.0\times 10^{-4}\: e^{-0.2x}\: \: \hat{a_{y}}   T ,Find the power required to move the conductor at constant speed to x=2.0 m,

y = 0 m in 5\times 10^{-3}s.Assume parallel motion along the x-axis.

A uniform magnetic field  \mathrm{\overrightarrow{B}=(3 \hat{i}+4 \hat{j}+\hat{k})}  exists in region of space. A semicircular wire of radius 1 m carrying current 1 A having its centre at (2,2,0) is placed in x-y plane as shown in fig. The force on semicircular wire will be

In the figure shown a current \mathrm{I}_1 is established in the long straight wire AB. Another wire CD carrying current \mathrm{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 resultant force on the wire CD is:

Two parallel wires carry currents of 10 A and 40 A in opposite directions. Another wire carrying a current antiparallel to 10 A is placed midway between the two wires. The magnetic force on it will be.

Two long parallel wires separated by a distance R have equal current I flowing in each. The magnetic field of one exerts a force F on the other. The distance R is increased to 2 R and the current in each wire is reduced from I to I / 2. What is the force between them now?

A straight horizontal conducting rod of the length 0.5 \mathrm{~m}  and mass  50 \mathrm{~gm}  is suspended by two vertical wires at its end. A current of 5.0 A is set up in the rod through the wires. what magnetic field should be set up normally to the conductor so that the tension in the wires is zero? Ignore the mass of the wires and take  \mathrm{g=10 \mathrm{~m} / \mathrm{s}^{2} }

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A power line lies along the east-west direction and carries a current of 10 \mathrm{~A}  The force per meter due to the earth's magnetic field of  10^{-4} \mathrm{~T} is:

A rectangular loop PQRS carrying a current i is situated near a long straight wire AB. If a steady current I is passed through AB as shown in Fig., the loop will

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Two very long parallel wires, separated by a distance d, carry equal current I in the same direction. At a certain instant of time, a point charge q is at a point P which is equidistant from the two wires, in the plane containing the two wires. If v is the velocity of the charge at this instant is perpendicular to this plane, the force due to the magnetic field at P will.

Concepts Covered - 1

Magnetic force on a current carrying conductor

Magnetic force on a current carrying conductor - 

In case of current carrying conductor in a magnetic field force experienced by its small length element is $d \vec{F}=i(d \vec{l} \times \vec{B})$

                                                                     
For total force, we will integrate the above equation. So the total magnetic force -

                                                   

$
\vec{F}=\int d \vec{F}=\int i(d \vec{l} \times \vec{B})
$


If magnetic field is uniform i.e., $\vec{B}=$ constant and $\int d \vec{l}=\vec{L}=$ vector sum of all the length elements from initial to final point. Which is in accordance with the law of vector addition is equal to length vector $\bar{L}^{\prime}$ joining initial to final point.

Then,

$
\vec{F}=i\left[\int d l\right] \times \vec{B}=i(\vec{L} \times \vec{B})
$
 

Direction of force - 

The direction of force is perpendicular to both the length and magnetic field vector as we have discussed earlier that the result of the cross product of two vector have direction perpendicular to both the vectors. It can be find by right hand palm rule, screw rule, right hand thumb rule etc. Here we will discuss one important rule for this i.e., Fleming’s left-hand rule.

According to Fleming’s left-hand rule - Stretch the fore-finger, central finger and thumb left hand mutually perpendicular. Then if the fore-finger points in the direction of field \vec{B}  and the central in the direction of current i, the thumb will point in the direction of force. For better understanding, look at the image given below, 

                                                                          

Note - If curved wire is given in the question then the length will be taken as shown in the figure - 

                                                                        

And the direction of the length vector should be in the direction of current.

 

 

 

 

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Magnetic force on a current carrying conductor

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