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  • Solve! 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 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 direction

  • Option 1)

     1/√2(−j^+k^)

     

  • Option 2)

    1/√3(−j^+k^+i^)
     

  • Option 3)

    1/√3(i^+j^+k^)
     

  • Option 4)

     1/√2(i^+k^)

 

Answers (1)

As we learnt in

Magnetic Field due to Circular Current at the centre -

If a coil of radius r is carrying current i then magnetic feild at a distnace X from its centre is:

B_{centre}= \frac{\mu_{0} }{4\pi } \frac{2\pi Ni}{r} = \frac{\mu_{0} Ni}{2r}

- wherein

Magnetic field due to horizontal loop is in \hat{k} and magnetic field due to vertical loop is along \hat{i} . So the resultant field is \vec{B}=B_o\hat{i}+B_o\hat{k}

Resultant Direction is  \frac{\vec{B}}{\left | \vec{B} \right |}=\frac{1}{\sqrt{2}}(\vec{i}+\vec{k})


Option 1)

 1/√2(−j^+k^)

 

incorrect

Option 2)

1/√3(−j^+k^+i^)
 

incorrect

Option 3)

1/√3(i^+j^+k^)
 

incorrect

Option 4)

 1/√2(i^+k^)

correct 

Posted by

Vakul

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