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

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. $B\propto I.\mathrm{B}$ is inversely proportional to the distance 'r' from the /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:

$$B=\frac{{\mu }_0}{4\pi }\cdot \frac{i}{r}(\sin {\varphi }_1+\sin {\varphi }_2)$$

From figure, $\alpha =({90}^{\circ }-{\varphi }_1)$ and $\beta =({90}^{\circ }+{\varphi }_2)$

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

$B=\frac{{\mu }_0}{4\pi }\cdot \frac{i}{r}(2\sin \varphi )$

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

$B=\frac{{\mu }_0}{4\pi }\frac{i}{r}[\sin {90}^{\circ }+\sin {90}^{\circ }]=\frac{{\mu }_0}{4\pi }\frac{2i}{r}$

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

$B=\frac{{\mu }_0}{4\pi }\frac{i}{r}[\sin {90}^{\circ }+\sin 0^{\circ }]=\frac{{\mu }_0}{4\pi }\frac{i}{r}$

• 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 the magnetic field produced by straight wire carrying current is obtained by maxwell's right hand thumb rule.