1 Crore+ students have signed up on EduRev. Have you? 
Determine the magnetic field at the centre of the current carrying wire arrangement shown in the figure. The arrangement extends to infinity. (The wires joining the successive squares are along the line passing through the centre)
Figure shows an amperian path ABCDA. Part ABC is in vertical plane PSTU while part CDA is in horizontal plane PQRS. Direction of circulation along the path is shown by an arrow near point B and at D. for this path according to Ampere’s law will be :
In the figure shown A B C D E F A was a square loop of side l, but is folded in two equal parts so that half of it lies in x z plane and the other half lies in the i'. The magnetic field at the centre is :
Due to FABC the magnetic field at O is along yaxis and due to CDEF the magnetic field is along xaxis.
Hence the field will be of the form
Calculating field due to FABC :
due to AB :
Due to BC :
Similarly due to FA :
Similarly due to CDEF :
A particle is moving with velocity and it produces an electric field at a point given by It will produce magnetic field at that point equal to (all quantities are S.I. units)
A ring of mass m, radius r having charge q uniformly distributed over it and free to rotate about its own axis is placed in a region having a magnetic field B parallel to its axis. If the magnetic field is suddenly switched off, the angular velocity acquired by the ring is :
If the magnetic field at ‘P’ in the given figure can be written as then K is:
Let us compute the magnetic field due to any one segment :
Resultant field will be :
Two infinitely long linear conductors are arranged perpendicular to each other and are in mutually perpendicular planes as shown in figure. If I_{1} = 2A along the yaxis and I_{2} = 3A along negative zaxis and AP = AB = 1cm. The value of magnetic field strength at P is 
Magnetic field strength at P due to I_{1}
Magnetic field strength at P due the I_{2
}
An infinitely long wire carrying current I is along Y axis such that its one end is at point A(0, b) while the wire extends upto +∞. The magnitude of magnetic field strength at point (a, 0) is.
Two long cylinders (with axis parallel) are arranged as shown to form overlapping cylinders, each of radius r, whose centers are separated by a distance d. Current of density J (Current per unit area) flows into the plane of page along the right shaded part of one cylinder and an equal current flows out of the plane of the page along the left shaded part of the other, as shown in the figure. The magnitude and direction of magnetic field at point O (O is the origin of shown xy axes) are :
Let the current density in complete left cylinder is , then current density in complete right cylinder is . Then magnetic field at any point P in the region of overlap is
Therefore magnitude of field at any point in region of overlap is and its direction is along positive ydirection at any point P in overlap region.
The correct answer is: in the + y direction
The magnetic field at the origin due to the current flowing in the wire as shown in figure below is :
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 








