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Two Long, Straight Wires, Each Carrying a Current of 5 A, Are Placed Along the X- and Y-axis Respectively. the Currents Point Along the Positive Directions of the Axes.

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प्रश्न

Two long, straight wires, each carrying a current of 5 A, are placed along the x- and y-axis respectively. The currents point along the positive directions of the axes. Find the magnetic fields at the points (a) (1 m, 1 m), (b) (−1 m, 1 m), (c) (−1 m, −1 m) and (d) (1 m, −1 m). 

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उत्तर

Given:
Magnitude of current, I = 5 A
Separation of the point from the wire, d = 1 m 
Thus, the magnitude of magnetic field due to current in the wires is given by 

\[B_1  =  B_2  = \frac{\mu_0 I}{2\pi d}\]

(a) At point (1 m, 1 m), the magnetic fields due to the wires are the same in magnitude, but they are opposite in direction.
Hence, the net magnetic field is zero.

(b) At point (−1 m, 1 m), the magnetic fields due to the wires are in upward direction.

\[\Rightarrow  B_{net}    =    B_1  +  B_2 \] 

\[                             =   \left( \frac{2 \times {10}^{- 7} \times 5}{1} + \frac{2 \times {10}^{- 7} \times 5}{1} \right)\] 

  = 2 × 10−6 T      (Along the z-axis)

(c) At point (−1 m, −1 m), the magnetic fields due to the wires are the same in magnitude, but they are opposite in direction.
Hence, the net magnetic field is zero.

(d) At point (1 m, −1 m), the magnetic fields due to the wires are in upward direction.

\[\Rightarrow  B_{net}    =    B_1  +  B_2 \] 

\[                             =   \left( \frac{2 \times {10}^{- 7} \times 5}{1} + \frac{2 \times {10}^{- 7} \times 5}{1} \right)\] 

               = 2 × 10−6 T    (Along the negative z-axis)

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अध्याय 35: Magnetic Field due to a Current - Exercises [पृष्ठ २५०]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 35 Magnetic Field due to a Current
Exercises | Q 11 | पृष्ठ २५०

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