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

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

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

  • 35 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           

 

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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

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A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B', at radial distances \frac{\text{a}} {2} and 2a respectively, from the axis of the wire is:

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Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that 'O' is their common point for the two. The wires carry I1 and I2 currents, respectively. Point 'P' is lying at distance 'd'  from 'O' along a direction perpendicular to the plane containing the wires. The magnetic field at the point 'P' will be :

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Concepts Covered - 1

Magnetic Field due to current in straight wire

Magnetic Field due to current in straight wire:

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

Magnitude of magnetic field B, produced by straight current carrying wire at a given point  is directly  proportional to current I pairing through the wire i.e.   B\propto I . B is inversely proportional to the distance 'r' from wire.

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} \cdot \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,    B=\frac{\mu_{o}}{4 \pi} \cdot \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 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}

 

  • When point P lies on axial position of current carrying conductor then magnetic field at P, B=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 direction of current in straight wire the known then direction of magnetic field produced by 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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