NEET-XII-Physics
36: Permanent Magnets
- #13A very long bar magnet is placed with its north pole coinciding with the centre of a circular loop carrying as electric current i. The magnetic field due to the magnet at a point on the periphery of the wire is B. The radius of the loop is a. The force on the wire is
(a) very nearly 2Ï€aiB perpendicular to the plane of the wire
(b) 2Ï€aiB in the plane of the wire
(c) πaiB along the magnet
(d) zerodigAnsr: aAns : (a) very nearly 2`` \,\mathrm{\,\pi \,}``aiB perpendicular to the plane of the wire
In this case, the north pole of the magnet is coinciding with the centre of the circular loop carrying electric current i. So, the magnetic field lines almost lie on the plane of the ring and the force due to the field lines is perpendicular to the field lines and to the plane of the circular ring.
Let idl be the current element, B be the magnetic field and dF be the force on the current element idl.
Now
dF = Bidl `` \Rightarrow F={\int }_{0}^{2\,\mathrm{\,\pi a\,}}\,\mathrm{\,B\,}idl``
`` \Rightarrow F=2\,\mathrm{\,\pi a\,}i\,\mathrm{\,B\,}``
Thus, the force acting on the wire is 2`` \,\mathrm{\,\pi \,}``aiB and it is perpendicular to the plane of the wire.
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