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

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. 

                                   

 

Derivation:

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 \$ivarphi\$, we determine $đ y, \mathrm{x}$ and ltheta 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 \left(-\phi_1\right)\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, $\alpha=\left(90^{\circ}-\phi_1\right)$ 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,

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

Note: 

• 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.