(a) Derive the expression for the torque acting on the rectangular current-carrying coil of a galvanometer. Why is the magnetic field made radial? (b) An α-particle is accelerated through a potential difference of 10kV and moves along the x-axis. It enters a region of uniform magnetic field B=2×10−3T acting along the y-axis. Find the radius of its path. (Take mass of α-particle =6.4×10−27kg.)
Answer: (a) τ=NIAB (radial field makes it position-independent). (b) r=10m.
Solution
(a) The forces on the two vertical sides (length ℓ) are BIℓ each, forming a couple with arm b: τ=BIℓb⋅N=NIAB (with A=ℓb). The field is made \textbf{radial} so that the coil's plane is always parallel to B; then sinθ=1 always and τ=NIAB is independent of the coil's position, giving a uniform (linear) scale. (b) Energy gained: K=qV=(2e)(104)=(3.2×10−19)(104)=3.2×10−15J. v=m2K=6.4×10−272(3.2×10−15)=1.0×106m/s.
