Question

A square loop of uniform conducting wire is as shown in figure. A
current-I (in amperes) enters the loop from one end and leaves the loop from opposite end as shown in figure. The length of one side of square loop is l metre. The wire has uniform cross section area and uniform linear mass density. In four situations of Column-I, the loop is subjected to four different uniform and constant magnetic field. Under the conditions of column I, match the column-I with corresponding results of column-II. ( $B_0$ in column I is a positive non-zero constant).



$\begin{array}{cc}ColumnI& ColumnII\\ \left(A\right)\overrightarrow{B}=B_0\hat{i}intelsa& \left(p\right)magnitudeofnetforceonloopis\sqrt{2}{B}_{0}Ilnewton\\ \left(B\right)\overrightarrow{B}=B_0\hat{i}intelsa& \left(Q\right)magnitudeofnetforceonloopiszero\\ \left(C\right)\overrightarrow{B}=B_0(\hat{i}+\hat{j})intesla& \left(R\right)magnitudeofnettorqueonloopaboutitscentreiszero\\ \left(D\right)\overrightarrow{B}=B_0\hat{k}intesla& \left(S\right)Netforceperpendiculartotheplaneofloop\end{array}$

• A. A-Q,R B-Q,R C-Q,R D-P,R
• B. A-R B-Q,S C-P D-P,R
• C. A-Q B-Q,R C-S D-Q
• D. A-Q B-Q C-Q,R D-P

Answer 1

The correct option is A A-Q,R B-Q,R C-Q,R D-P,R

(A) Because the magnetic field is parallel to x- axis, the force on wire parallel to x- axis is zero. The force on each wire parallel to y-axis is $B_0\frac{1}{2}l$. Hence net force on loop is $B_{0}Il$, $\perp$ to loop plane (downward). Since force on each wire parallel to y- axis passes through centre of the loop net torque about centre of the loop is zero.

(B) Because the magnetic field is parallel to y- axis, the force on wire parallel to y- axis is $B_0\frac{1}{2}l$. Hence net force on loop is $B_{0}Il$, $\perp$ to loop plane (upward). Since force on each wire parallel to x- axis passes through centre of the loop, net torque about centre of the loop is zero.

(C) Since net displacement of current from entry point in the loop to exit point in the loop is along the diagonal of the loop. The direction of external uniform magnetic field is also along the same diagonal. Hence net force on the loop is zero. Since force on each wire on the loop passes through centre of the loop net torque about centre of the loop is zero.

(D) The net displacement of current from entry point in the loop to exit point in the loop is along the diagonal (of length $\sqrt{2l}$ ) of the loop. The direction of external uniform magnetic field is also perpendicular to the same diagonal. Hence magnitude of net force on the loop is $B_{0}I\sqrt{2l}$ Since force on each wire on the loop passes through centre of the loop net torque about centre of the loop is zero.