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Karnataka Board PUCPUC Science Class 11

Four Long, Straight Wires, Each Carrying a Current of 5.0 A, Are Placed in a Plane as Shown in Figure. the Points of Intersection Form a Square of Side 5.0 Cm. - Physics

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Question

Four long, straight wires, each carrying a current of 5.0 A, are placed in a plane as shown in figure. The points of intersection form a square of side 5.0 cm.
(a) Find the magnetic field at the centre P of the square.
(b) Q1, Q2, Q3, and Q4, are points situated on the diagonals of the square and at a distance from P that is equal to the diagonal of the square. Find the magnetic fields at these points. 

Short/Brief Note
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Solution

Given: 

Let the horizontal wires placed at the bottom and top are denoted as W1 and W2 respectively.
Let the vertical wires placed at the right and left to point P are denoted as W3 and W4 respectively.
Magnitude of current, I = 5 A
(a) Consider point P.
Magnetic fields due to wires W1 and W2 are the same in magnitude, but they are opposite in direction.
Magnetic fields due to wires W3 and W4 are the same in magnitude, but they are opposite in direction.
Hence, the net magnetic field is zero.
Net magnetic field at P due to these four wires = 0

(b) Consider point Q1.
Due to wire W1, separation of point Q1 from the wire (d) is 7.5 cm.
So, the magnetic field due to current in the wire is given by

\[B_{W_1}  = \frac{\mu_0 I}{2\pi d}\]

       = 4 × 10−5 T     (In upward direction)

Due to wire W2, separation of point Q1 from the wire (d) is 2.5 cm.
So, the magnetic field due to current in the wire is given by

\[B_{W_2}  = \frac{4}{3} \times  {10}^{- 5}   T\]    (In upward direction)

Due to wire W3, separation of point Q1 from the wire (d) is 7.5 cm.
So, the magnetic field due to current in the wire is given by
BW3 = 4 × 10−5 T   (In upward direction)

Due to wire W4, separation of point Q1 from the wire (d) is 2.5 cm.
So, the magnetic field due to current in the wire is given by

\[B_{W_4}  = \frac{4}{3} \times  {10}^{- 5}   T\]   (In upward direction)

∴ Net magnetic field at point Q1

\[B_{Q_1}  = \left( 4 + \frac{4}{3} + 4 + \frac{4}{3} \right) \times  {10}^{- 5}   \] 

\[             = \frac{32}{3} \times  {10}^{- 5} \] 

\[             = 1 . 06 \times  {10}^{- 4}   T        (\text{ In  upward  direction })\]

At point Q2,
Magnetic field due to wire W1:
BW1 = 4 × 10−5 T   (In upward direction)
Magnetic field due to wire W2:
\[B_{W_2}  = \frac{4}{3} \times  {10}^{- 5}   T\]    (In upward direction)
Magnetic field due to wire W3:
\[B_{W_3}  = \frac{4}{3} \times  {10}^{- 5}   T\]   (In downward direction)
Magnetic field due to wire W4:
\[B_{W_4}  = 4 \times  {10}^{- 5}   T\]      (In downward direction)
∴ Net magnetic field at point Q2, \[B_{Q_2}  = 0\]
Similarly, at point Q3, the magnetic field is 1.1 × 10−4 T (in downward direction) and at point Q4, the magnetic field is zero.
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Chapter 13: Magnetic Field due to a Current - Exercises [Page 250]

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HC Verma Concepts of Physics Vol. 2 [English] Class 11 and 12
Chapter 13 Magnetic Field due to a Current
Exercises | Q 12 | Page 250

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