A non-planar loop of conducting wire carrying a current $I$ is placed as shown in the figure. Each of the straight sections of the loop is of length $2 a$ . The magnetic field due to this loop at the point $P (a, 0, a)$ points in the direction [2001]

Question

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(4)

Here loop $A D E F A$ lies in the $y - z$ plane and loop $A B C D A$ lies in the $x - y$ plane.

By choosing the loops, we find that in one loop we have to take current from $A$ to $D$ and in the other one from $D$ to $A$ . Effectively there is no current in $A D$ . Hence these two currents cancel out the effect of each other as far as creating magnetic field at the point $P$ is considered.

The point $(a, 0, a)$ lies in the $X - Z$ plane.

The magnetic field due to current in $A B C D A$ will be in the positive $Z$ -direction.

Due to symmetry, the $y$ -components and $x$ -components will cancel out each other.

Similarly, the magnetic field due to current in $A D E F A$ will be in the $x$ -direction.

$∴$ The direction of resultant magnetic field at $P (a, 0, a)$ $\vec{B} = \frac{1}{\sqrt{2}} (\hat{i} + \hat{k})$ .

Solution

Q.
A non-planar loop of conducting wire carrying a current $I$ is placed as shown in the figure. Each of the straight sections of the loop is of length $2 a$ . The magnetic field due to this loop at the point $P (a, 0, a)$ points in the direction [2001]

1 $\frac{1}{\sqrt{2}} (- \hat{j} + \hat{k})$

2 $\frac{1}{\sqrt{3}} (- \hat{j} + \hat{k} + \hat{i})$

3 $\frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k})$

4 $\frac{1}{\sqrt{2}} (\hat{i} + \hat{k})$

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Solution

Q. A non-planar loop of conducting wire carrying a current I is placed as shown in the figure. Each of the straight sections of the loop is of length 2a. The magnetic field due to this loop at the point P(a,0,a) points in the direction [2001]

1 12(-j^+k^)

2 13(-j^+k^+i^)

3 13(i^+j^+k^)