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Force on a wire of length $\vec{l}$ carrying a current $I$ placed in a uniform magnetic field $\vec{B}$ is given by \begin{align} \vec{F}=I\;\vec{l}\times\vec{B} \end{align}

**Problem (IIT JEE 2015): **
A conductor (shown in the figure) carrying constant current $I$ is kept in the $x\text{-}y$ plane in a uniform magnetic field $\vec{B}$. If $F$ is the magnitude of the total magnetic force acting on the conductor, then the correct statement(s) is (are),

- If $\vec{B}$ is along $\hat{z}$, $F\propto(L+R)$
- If $\vec{B}$ is along $\hat{x}$, $F=0$
- If $\vec{B}$ is along $\hat{y}$, $F\propto(L+R)$
- If $\vec{B}$ is along $\hat{z}$, $F=0$

**Solution: **

Case (A): \begin{align} \vec{F} & =I(2(L+R)\,\hat{x})\times(B\hat{z}) \\ &=-2IB(L+R)\,\hat{y}, \end{align} Case (B): \begin{align} \vec{F} & =I(2(L+R)\,\hat{x})\times(B\hat{x})=0, \end{align} Case (C): \begin{align} \vec{F} &=I(2(L+R)\,\hat{x})\times(B\hat{y}) \\ &=2IB(L+R)\,\hat{z}. \end{align} Case (D): \begin{align} \vec{F} &=I(2(R+L)\,\hat{x})\times(B\hat{z}) \\ &=-2IB(R+L)\,\hat{y} \end{align} Thus, correct options are A, B, and C.

**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{align} \vec{F}_\text{AB} & =\int_{A}^{B} I_2\,\mathrm{d}\vec{l}\times\vec{B}=0.\nonumber \end{align} 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 \begin{align} \vec{F}_\text{BC}=F_\text{BC}\odot, \end{align} 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 \begin{align} \vec{F}_\text{DA}=F_\text{DA}\otimes, \end{align} where $F_\text{DA}$ is the magnitude of force. By symmetry, \begin{align} \vec{F}_\text{BC}=-\vec{F}_\text{DA}. \end{align}

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{align} 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{align} Thus, correct options are A and C.

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