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1) A non planar loop of conducting wire carrying a current I is placed as shown .Each of the straight section of the loop have a length = 2a. The magnetic field due to this loop at the point P (a,0,a) points in the direction --- A) 1/√2*(-j + k) B) 1/√3*(-j + k +i) C) 1/√3*(i + j + k) D) 1/√2*(i + k) (i, j, k represent the respective Unit vectors) 2) A current of 1A is passed thro' a straight wire of length 2mt. The magnetic field at a point in air at a distance of 3mt from either side of the wire & lying on the axis of wire will be ?


1) A non planar loop of conducting wire carrying a current I is placed as shown .Each of the straight section of the loop have a length = 2a. The magnetic field due to this loop at the point P (a,0,a) points in the direction ---

 

A) 1/√2*(-j + k)                 B) 1/√3*(-j + k +i)                   C) 1/√3*(i + j + k)                D) 1/√2*(i + k)         

(i, j, k represent the respective Unit vectors)

 

 

2) A current of 1A is passed thro' a straight wire of length 2mt. The magnetic field at a point in air at a distance of 3mt from either side of the wire & lying on the axis of wire will be ?


Grade:upto college level

1 Answers

Rohit Ashiwal
28 Points
4 years ago
ans 1) If you apply right hand thumb rule you shall find that resultant magnetic field is a superposition of 2 field in direction of x-axis and z-axis only hence D will be the correct choice as it was asked to find a unit vector.ans 2) For such kind of systems we use this formula :-B = (u/4πr)*(sina + sinb) (direction by thumb rule)here a is angle made by line joining our point foot of perpendicular from that point and lower end, b is same angle for upper end; r is length of foot of perpendicular.Ergo required magnetic field is :-B = √8/12 * 10^(-7) T.

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