1.

There is one infinitely long strip of current with a large width , carrying current K per unit width. There is onelong solenoid which carries current I. The solenoid is kept near the sheet with its axis parallel to the width of sheet and it is found that the magnetic field near the centre of the solenoid is zero. If the solenoid is further rotated in such a manner that its length is perpendicular to the sheet and current and one end of the solenoid is near the sheet, what will be the net magnetic field at the centre of this end of solenoid?

Answer»

`(mu_(0)Ksqrt5)/4` , parallel to the strip
`(mu_(0)K sqrt5)/4` , at an angle `tan^(-1)` (1/2) with the plane of strip
`(mu_(0)Ksqrt5)/4 `, at an angle `tan^(-1) (2)` with the plane of strip
`(mu_(0)Ksqrt2)/4` , parallel to the strip

Solution :We should be aware that the end of a LONG solenoid magnetic FIELD is half of the value near its centre. Moreover, length of solenoid is perpendicular to sheet, so these two fields are perpendicular. The RESULTANT field will be INCLINED at an angle `theta` with the plane of strip.
`B_("net") = sqrt(((mu_(0)K)/2)^(2) + ((mu_(0)K)/4)^(2)) = (mu_(0)K)/4 sqrt(4 + 1) = (mu_(0)Ksqrt5)/4`
`tan theta = 4/((mu_(0)K)/2) = 1/2 rArr theta = tan^(-1) (1//2)`


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