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 rotated by a 90^(@) angle from theprevious placement, in such a manner that its axis is parallel to the direction of current, what will be the net magnetic field near the centre of the solenoid?

There is one infinitely long strip of current with a large width , car

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 rotated by a 90^(@) angle from theprevious placement, in such a manner that its axis is parallel to the direction of current, what will be the net magnetic field near the centre of the solenoid?
Answer» `(mu_(0)K)/2 sqrt2`, perpendicular to the PLANE of strip `mu_(0)K sqrt2 `, parallel to the plane of strip `(mu_(0)K)/2 sqrt2`, parallel to the plane of strip `(mu_(0)K)/2 sqrt5`, parallel to the plane of strip Solution :Now we know that magnetic field near the CENTRE of SOLENOID due to its own current is also `(mu_(0)K)/2`,which is same as that due to the strip of current. When the solenoid is ROTATED by `90^(@)`in such a MANNER that its axis is parallel to current, then its magnetic field becomes perpendicular to the magnetic field of strip but both the fields are still parallel to the plane of strip. Thus, the net magnetic field will be `(mu_(0)K)/2 sqrt2`parallel to the plane of strip.

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