# Magnetic Field Near a Current Carrying Straight Wire

## Problems from IIT JEE

**Problem (IIT JEE 2000): **
An infinitely long conductor PQR is bent to form a right angle as shown in figure. A current $I$ flows through PQR. The magnetic field due to this current at the point M is $B_1$. Now, another infinitely long straight conductor QS is connected at Q, so that current is $I/2$ in QR as well as in QS, the current in PQ remaining unchanged. The magnetic field at M is now $B_2$. The ratio $B_1/B_2$ is given by

- 1/2
- 1
- 2/3
- 2

**Solution: **
The magnetic field at the point P (see figure) by a current carrying wire is given by
\begin{alignat}{2}
&B=\frac{\mu_0I}{4\pi d}(\cos\theta_1-\cos\theta_2).\nonumber
\end{alignat}
In first case, the field at the point M by part PQ, part QR, and total field are given by
\begin{alignat}{2}
&B_\text{PQ}=\frac{\mu_0I}{4\pi d}(\cos 0-\cos 90)=\frac{\mu_0I}{4\pi d}, \nonumber\\
&B_\text{QR}=\frac{\mu_0I}{4\pi d}(\cos 180-\cos 180)=0,\nonumber\\
&B_1=B_\text{PQ}+B_\text{QR}=\frac{\mu_0I}{4\pi d}. \nonumber
\end{alignat}
In second case, the field at M by part PQ and part QR remains same as in first case. The field by part QS and total field are given by,
\begin{alignat}{2}
&B_\text{QS}=\frac{\mu_0(I/2)}{4\pi d}(\cos90-\cos 180)=\frac{\mu_0I}{8\pi d}, \nonumber \\
&B_2=B_\text{PQ}+B_\text{QR}+B_\text{QS}=\frac{3}{2}\frac{\mu_0I}{4\pi d}. \nonumber
\end{alignat}