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

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

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

  • 27 Questions around this concept.

Solve by difficulty

 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 centre 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 rod PQ is connected to the capacitor plates. The rod is placed in a magnetic field (B) directed downward perpendicular to the plane of the paper. If the rod is pulled out of magnetic field with velocity v as shown in figure

A very long current carrying wire is placed along z-axis having current of magnitude i1 towards negative z-axis. A semicircular wire of radius R and having current i2 is placed in x-y plane, such that line joining two end points of the semicircular wire passes through long wire as shown in fig. Nearest distance of semicircular wire from long wire is R. The net magnetic force on semicircular wire will be

A wire abc is carrying current I. It is bent as shown in figure placed in a uniform magnetic field of induction B . Length ab=l and \angleabc = 45 °. The ratio of force on ab and on bc is

An infinitely long, straight conductor AB is fixed and a current is passed through it. Another movable straight wire CD of finite length and carrying current is held perpendicular to it and released. Neglect weight of the wire

A hypothetical magnetic fields existing in a region is given by  \overline{\mathrm{B}}=\mathrm{B}_0 \hat{\mathrm{e}}_{\mathrm{r}}  where \hat{\mathrm{e}}_{\mathrm{r}}  denotes the unit vector along the radial direction. A circular loop of radius ' a ' carrying a current I, is placed with its plane parallel to the x-y plane and the centre at (0,0, d). The magnitude of the magnetic force acting on the loop is

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A straight wire of length 0.5 metre and carrying a current of 1.2 ampere is placed in uniform magnetic field of induction 2 tesla. The magnetic field is perpendicular to the length of the wire. The force on the wire is

A current carrying coil is subjected to a uniform magnetic field. The coil will orient so that its plane becomes.

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A square loop ABCD carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be :

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 
\begin{array}{l}{\int d \vec{l}=\vec{L}=\text { vector sum of all the length elements from initial to final point. Which is in accordance with }} \\ {\text { the law of vector addition is equal to length vector } \bar{L}^{\prime} \text { joining initial to final point. }}\end{array}

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

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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