Magnetics 3 Question 13

13. Two long straight parallel wires are 2m apart, perpendicular to the plane of the paper.

The wire A carries a current of 9.6A, directed into the plane of the paper. The wire B carries a current such that the magnetic field of induction at the point P, at a distance of 10/11m from the wire B, is zero.

(1987,7M) Find

(a) the magnitude and direction of the current in B.

(b) the magnitude of the magnetic field of induction at the point S.

(c) the force per unit length on the wire B.

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Solution:

  1. (a) Direction of current at B should be perpendicular to paper outwards. Let current in this wire be iB. Then,

μ02πiA2+1011=μ02πiB(10/11)

or iBiA=1032

or iB=1032×iA=1032×9.6=3A

(b) Since, AS2+BS2=AB2

ASB=90

At S:B1= Magnetic field due to iA

=μ02πiA1.6=(2×107)(9.6)1.6=12×107T

B2= Magnetic field due to iB

=μ02πiB1.2=(2×107)(3)1.2=5×107T

Since, B1 and B2 are mutually perpendicular. Net magnetic field at S would be

B=B12+B22=(12×107)2+(5×107)2=13×107T

(c) Force per unit length on wire B

Fl=μ02πiAiBr(r=AB=2m)=(2×107)(9.6×3)2=2.88×106N/m



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