Force on a current carrying conductor moving in uniform magnatic field (By pak physics notes)

Consider a conductor of length "L", area of cross section "a" placed in a uniform magnetic field of "B".
Let there are "n" number of charges (e) per unit volume and an electric current "I" is passing through it.

	Volume of conductor = V = area x length = a x L Number of electrons = N = n . Volume Charge on each electron = e Total charge = q= Ne We know that the force experienced by a charge in a uniform magnetic field is
	F = Q (V x B )
	Putting the value of "q" F = Ne (vBSinq) F = nV. e(vBSinq) F = ne . aL (vBSinq) F = naLe (v B Sinq)
	Now consider the length of conductor (L) as a vector quantity in the direction of velocity vector.
	F = nave(Lv B Sinq) F = nave(LB Sinq)
	Since S = vt v = s/t v = L/t
	Putting the value of v
	F = na(L/t)e (L B Sinq) F = naLe/t (L B Sinq)
	Since naLe = q Therefore,
	F = q/t (L B Sinq)
	Also q/t = I
	F = I (L B Sinq)
	F = ILB Sinq
	The expression clearly indicates that the force acting on the wire when placed in a uniform magnetic field is perpendicular to the plane formed by the L and B. i.e. F is perpendicular to L and also perpendicular to B.

                        (www.pak-physicsnotes.blogspot.com)

If You Enjoyed This Post Please Take a Second To Share It.

You Might Also Like

Stay Connected With Free Updates

Subscribe via Email