# Effect of a Uniform Magnetic Field on a Current Loop

## Problems from IIT JEE

**Problem (IIT JEE 2006): **
An infinitely long wire carrying current $I_1$ passes through O and is perpendicular to the plane of paper. Another current carrying loop ABCD lies in plane of paper as shown in figure. Which of the following statement(s) is (are) correct?

- net force on the loop is zero.
- net torque on the loop is zero.
- loop will rotate clockwise about axis $OO^\prime$ when seen from $O$.
- loop will rotate anticlockwise $OO^\prime$ when seen from $O$.

**Solution: **
The magnetic field $\vec{B}$ by the current $I_1$ is circumferential. For each element $\mathrm{d}\vec{l}$ on the branch AB, $\vec{B}$ is parallel to the current direction making the force,
\begin{alignat}{2}
&\vec{F}_\text{AB}=\int_{A}^{B} I_2\,\mathrm{d}\vec{l}\times\vec{B}=0.\nonumber
\end{alignat}
For the branch CD, $\vec{B}$ is anti-parallel to $\mathrm{d}\vec{l}$. Thus, total force on branch CD is $\vec{F}_\text{CD}=0$.

For the branch BC, $\vec{B}$ is perpendicular to $\mathrm{d}\vec{l}$, and $\mathrm{d}\vec{l}\times \vec{B}$ is coming out of the paper making $\vec{F}_\text{BC}=F_\text{BC}\odot$, where $F_\text{BC}$ is the magnitude of force. For the branch DA, $\mathrm{d}\vec{l}\times \vec{B}$ is going into the paper making $\vec{F}_\text{DA}=F_\text{DA}\otimes$, where $F_\text{DA}$ is the magnitude of force. By symmetry, $\vec{F}_\text{BC}=-\vec{F}_\text{DA}$.

Thus, net force acting on ABCDA is zero. However, there is non-zero clockwise (when looking from O) torque that rotates the loop in clockwise direction. The readers are encouraged to show that,
\begin{alignat}{2}
&F_\text{BC}=\int_{r_1}^{r_2} I_2 B(r) \mathrm{d} r=\frac{\mu_0 I_1 I_2}{2\pi}\ln\left(\frac{r_2}{r_1}\right). \nonumber
\end{alignat}
Thus, correct options are A and C.