Q. 123.7( 3 Votes )

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 from 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 diagonal of the square and at a distance from P that to equal to the length of the diagonal of the square. Find the magnetic fields at these points.



Answer :


Given:
Current in the 4 wires: i = 5.0 A
Side of the square : d = 5 cm = 0.05 m

In the diagram above, l1 and l2 represents vertical wires. l3 and l4 represent horizontal wires.
Arrows represent the direction of current in the wire. We can determine the direction of magnetic field due to any of the wire using Right hand rule.
Formula used:
By Ampere’s Law for a current carrying wire is
Here, B is the magnitude of magnetic field, μ0 is the permeability of free space and μ0= 4π × 10-7 T mA-1 and d is the distance between the current carrying wire and the required point.
(a)
At point P, if we use Right hand rule for every wire.
Field due to l1 would be into the plane and field due to l2 would come out of the plane. Thus canceling out each other.
Similarly, field due to l3 would go into the plane and field due to l4 would come out, canceling each other.
Thus, net field at P is zero.
(b)
At Q1, By geometry:
Distance between wire l1 and Q1 : d1 = 0.025 m
Distance between wire l2 and Q1 : d2 = 0.075 m
Distance between wire l3 and Q1 : d3 = 0.025 m
Distance between wire l4 and Q1 : d3 = 0.075 m
Net magnetic field at Q1 due to wires l1,l2,l3,l4 is
BQ1 = Bl1+Bl2+Bl3+Bl4


BQ1 = 1.06 × 10-4 T
Hence, magnetic field at Q1 is 1.06 × 10-4 T and in upward direction.
At Q2, Similarly by geometry
Distance between wire l1 and Q2 : d1 = 0.075 m
Distance between wire l2 and Q2 : d2 = 0.025 m
Distance between wire l3 and Q2 : d3 = 0.025 m
Distance between wire l4 and Q2 : d3 = 0.075 m
Now, Net magnetic field at Q2 due to wires l1,l2,l3,l4 is
BQ2 = Bl1+Bl2+Bl3+Bl4


BQ2 = 0
Hence, magnetic field at Q2 is zero.
Negative sign for Bl3 and Bl4 as direction of magnetic field due to l1(going in) is opposite to l4(coming out) and l2(going in) is opposite to l39coming out).
At Q3, By geometry
Distance between wire l1 and Q3 : d1 = 0.075 m
Distance between wire l2 and Q3 : d2 = 0.025 m
Distance between wire l3 and Q3 : d3 = 0.075 m
Distance between wire l4 and Q3 : d3 = 0.025 m
Net magnetic field at Q3 due to wires l1,l2,l3,l4 is
BQ3 = Bl1+Bl2+Bl3+Bl4


BQ3 = 1.06 × 10-4 T
Hence, magnetic field at Q3 is 1.06 × 10-4 T and in downward direction.
At Q4,
Magnetic field at Q4 would be same as that of Q2 because at Q4, l1 and l4 have opposite fields, l2 and l3 have opposite fields. Thus canceling out each other and BQ4 is zero.
Hence, Magnetic field at Q1 and Q3 have same magnitude of
1.06 × 10-4 T but in opposite direction and magnetic field at Q2 and Q4 is zero.


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