Question

A wire frame $ABCDEF$ is kept as shown in figure. $I$ is the current in the wire frame.

• A. The direction of magnetic field at $‘O^{\prime }$ due to the wire frame will be along $\left(\frac{-\hat{i}-\hat{j}}{\sqrt{2}}\right)$
• B. The direction of magnetic field at $P(0,a,0)$ due to wire $CD$ and $EF$ will be along $(-\hat{j})$.
• C. The direction of magnetic field at $Q(a,0,0)$ due to wire $EF$ and $CD$ will be along $(-\hat{j})$.
• D. The direction of magnetic field at $P(0,a,0)$ due to wire $AF$ and $CB$ will be along $(-\hat{i})$.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A The direction of magnetic field at $‘O^{\prime }$ due to the wire frame will be along $\left(\frac{-\hat{i}-\hat{j}}{\sqrt{2}}\right)$
B The direction of magnetic field at $P(0,a,0)$ due to wire $CD$ and $EF$ will be along $(-\hat{j})$.
C The direction of magnetic field at $Q(a,0,0)$ due to wire $EF$ and $CD$ will be along $(-\hat{j})$.
D The direction of magnetic field at $P(0,a,0)$ due to wire $AF$ and $CB$ will be along $(-\hat{i})$.

(A) Direction of $B$ at $O$ is same as direction of resultant magnetic dipole moment of the current carrying loop i.e. will be along $\left(\frac{-\hat{i}-\hat{j}}{\sqrt{2}}\right)$

(B) Using biot-savart law, direction of $B$ at point $P$ due to CD and EF is shown in figure.
Hence net magnetic field is along negative y axis.
(C)Using right hand rule,
The direction of $B$ at $Q$ due to CD and EF is shown in figure,
Hence net magnetic field is along negative y axis.
(D) Using right hand rule,
The direction of $B$ at $P$ due to AF and CB is shown in figure,
Hence net magnetic field is along negative x axis.