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Magnetic Field Due To Current In Straight Wire - Practice Questions & MCQ

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

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  • Magnetic Field due to current in straight wire is considered one of the most asked concept.

  • 56 Questions around this concept.

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 The magnetic field at the origin due to the current flowing in the  wire as shown in figure below is           


Two identical conducting wires AOB and COD are placed at right angles to each other. The wire AOB carries an electric current I_{1} and COD carries a current I_{2} . The magnetic field on a point lying at a distance d from O, in a direction perpendicular to the plane of the wires AOB and COD, will be given by

Concepts Covered - 1

Magnetic Field due to current in straight wire

Magnetic Field due to current in a straight wire:

Magnetic field lines around a current-carrying straight wire are concentric circles whose centre lies on the wire.


The magnitude of magnetic field B, produced by a straight current-carrying wire at a given point is directly proportional to the current I pairing through the wire i.e. B is inversely proportional to the distance 'r' from the wire \left ( B\propto \frac{1}{r} \right )  as shown in the figure given below. 




The directions of magnetic fields due to all current elements are the same in the figure shown, we can integrate the expression of magnitude as given by Biot-Savart law for the small current element  dy as shown in the figure


                                                                          B=\int \mathrm{dB}=\frac{\mu_{0}}{4 \pi} \int \frac{I {d} y \sin \theta}{x^{2}}

In order to evaluate this integral in terms of angle $\varphi$, we determine đy, x and \theta in terms of perpendicular distance "r" (which is a constant for a given point) and angle " \phi". Here,

                                                                                     \begin{aligned} y &=r \tan \phi \\ d y &=r \sec ^{2} \phi d\phi \\ x&=r \sec \phi \\ \theta &=\frac{\pi}{2}-\phi \end{aligned}

Substituting in the integral, we have :
                                                     \Rightarrow B=\frac{\mu_{0}}{4 \pi} \int \frac{I r\sec ^{2} \phi d \phi \sin \left(\frac{\pi}{2}-\phi\right)}{r^{2} \sec ^{2} \phi}=\frac{\mu_{0}}{4 \pi} \int \frac{I \cos \phi d \phi}{r}

Taking out I and r out of the integral as they are constant :
                                                   \Rightarrow B=\frac{\mu_{0} I}{4 \pi r} \int \cos \phi d \phi

Integrating between angle  -\phi_{1}   and   \phi_{2} , we have

                                                :\begin{gathered} \\ \Rightarrow B=\frac{\mu_{0} I}{4 \pi r} \int_{-\phi_{1}}^{\phi_{2}} I \cos \phi d \phi \\ \Rightarrow B=\frac{\mu_{0} I}{4 \pi r}\left(\sin \phi_{2}-\sin (-\phi_{1})\right) \end{gathered}

Note: -\phi_1 is taken  because it is measured in the opposite sense of  \phi_2 with respect to the reference line ( negative x-axis here)

                                                 \Rightarrow B=\frac{\mu_{0} I}{4 \pi r}\left(\sin \phi_{2}+\sin \phi_{1}\right)

Magnetic field due to a current-carrying wire at a point P which lies at a perpendicular distance r from the wire, as shown, is given as: 

                                                       B=\frac{\mu_{0}}{4 \pi} \frac{i}{r}\left(\sin \phi_{1}+\sin \phi_{2}\right)   

From figure, \alpha=\left(90^{\circ}-\phi_{1}\right) \text { and } \beta=\left(90^{\circ}+\phi_{2}\right) 

Hence,  it can be also written as  B=\frac{\mu_{o}}{4 \pi} \frac{i}{r}(\cos \alpha-\cos \beta) 


Different cases: 

Case 1: When the linear conductor XY is of finite length and the point P lies on it's perpendicular bisector as shown


B=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{r}(2 \sin \phi)

Case 2: When the linear conductor XY is of infinite length and the point P lies near the centre of the conductor

B=\frac{\mu_{0}}{4 \pi} \frac{i}{r}\left[\sin 90^{\circ}+\sin 90^{\circ}\right]=\frac{\mu_{0}}{4 \pi} \frac{2 i}{r}

Case 3: When the linear conductor is of semi-infinite length and the point P lies near the end Y or X

B=\frac{\mu_{0}}{4 \pi} \frac{i}{r}\left[\sin 90^{\circ}+\sin 0^{\circ}\right]=\frac{\mu_{0}}{4 \pi} \frac{i}{r}

Case 4: When point P lies on the axial position of the current-carrying conductor then magnetic field at P,

B=\frac{\mu_{o}}{4 \pi} \frac{i}{r}(\cos \alpha-\cos \beta)=\frac{\mu_{o}}{4 \pi} \frac{i}{r}(\cos 0-\cos 0)=0


  • The value of magnetic field induction at a point, on the centre of separation of two linear parallel conductors carrying equal currents in the same direction, is zero.
  • If the direction of current in the straight wire the known then the direction of the magnetic field produced by a straight wire carrying current is obtained by maxwell's right-hand thumb rule.


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Magnetic Field due to current in straight wire

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