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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
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
Updated On: Jun 14, 2022
- $\frac{1}{\sqrt{2}}(-\widehat{j}+\widehat{k})$
- $\frac{1}{\sqrt{3}}(-\widehat{j}+\widehat{k}+\widehat{i})$
- $\frac{1}{\sqrt{3}}(\widehat{i}+\widehat{j}+\widehat{k})$
- $\frac{1}{\sqrt{2}}(\widehat{i}+\widehat{k})$
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The Correct Option is D
Solution and Explanation
The magnetic field at P(a, 0, a) due to the loop is equal to the vector sum of the magnetic fields produced by loops ABCDA and AFEBA as shown in the figure.
Magnetic field due to loop ABCDA will be along $\widehat{i}$ and due to loop AFEBA, along $\widehat{k}$ Magnitude of magnetic field due to both the loops will be equal. Therefore, direction of resultant magnetic field at P will be $\frac{1}{\sqrt{2}}(-\widehat{i}+\widehat{k})$
NOTE This is a common practice, when by assuming equal currents in opposite directions in an imaginary wire (here AB) loops are completed and solution becomes easy.
Magnetic field due to loop ABCDA will be along $\widehat{i}$ and due to loop AFEBA, along $\widehat{k}$ Magnitude of magnetic field due to both the loops will be equal. Therefore, direction of resultant magnetic field at P will be $\frac{1}{\sqrt{2}}(-\widehat{i}+\widehat{k})$
NOTE This is a common practice, when by assuming equal currents in opposite directions in an imaginary wire (here AB) loops are completed and solution becomes easy.
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