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Question

A long straight wire carrying a current of 100 A is placed perpendicular to a uniform magnetic field of strength 1.0×104 Tesla as shown. The resultant field strength at P in magnitude at distance 10 cm from O is

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A
2.0×104 T
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B
2.23×104 T
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C
1.73×104 T
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D
0
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Solution

The correct option is B 2.23×104 T
The magnetic field B, due to a long infinite straight current carrying (I) wire at a distance r is given by:
B=μ02π.Ir
Given: I=100A,r=10cm=0.1m
Hence, B=4π×1072π.1000.1=2×104T
Now, according to Right hand thumb rule, the direction of this magnetic field at point P will be inwards to the plane of paper. A magnetic field of 1.0×104T is already there and both the fields are perpendicular to each other.
Therefore resultant field strength at point P will be:
B=(2×104)2+(1.0×104)2=2.23×104T

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