(1) A straight conductor carries a current of 10A. An electron traveling with a speed of 2.5×10^6m/s parallel to the conductor at a distance of 1cm from the conductor experiences a force of
(a) 8×10^-17N (b) 8×10^-18N (c) 8×10^-19N (d) 4×10^-20N (e) 4×10^-18N
Force on the electron, F=qvB, where B= μoI/2πr with usual notations.
Note that the direction of the magnetic field produced by the conductor is perpendicular to the direction of motion of the electron since the magnetic field lines are in the form of concentric circles.
Therefore, F= (1.6×10^-19×2.5×10^6×4π×10^-7×10) / 2π×0.01 = 8×10^-17N.
(2) A galvanometer having resistance 3Ω is converted into an ammeter by connecting a 3Ω shunt resistance across it. In order to double the range of this ammeter, the additional shunt resistance to be connected across it is
(a) 3Ω (b) 2.5Ω (c) 2Ω (d) 1.5Ω (e) 6Ω
With usual notations, Ig G = (I-Ig)S. Since G = S =3Ω, Ig = I/2. If S′ is the additional shunt required, the effective shunt for doubling the range is SS′/(S+S′) . Therefore, when the range is doubled, IgG = (2I-Ig) SS′/(S+S′) . Substituting G=S=3 and Ig= I/2 , 3S'/(3+S') =1. Therefore,S'=1.5Ω.
Instead of following the above mathematical steps, you can easily arrive at the answer as follows:
The ammeter with the original shunt has an effective resistance of 1.5Ω. If you want to pass twice the maximum allowed current into the ammeter, half of this doubled current has to be diverted through another path. So you have to provide a shunt path of resistance 1.5Ω.
(a) 8×10^-17N (b) 8×10^-18N (c) 8×10^-19N (d) 4×10^-20N (e) 4×10^-18N
Force on the electron, F=qvB, where B= μoI/2πr with usual notations.
Note that the direction of the magnetic field produced by the conductor is perpendicular to the direction of motion of the electron since the magnetic field lines are in the form of concentric circles.
Therefore, F= (1.6×10^-19×2.5×10^6×4π×10^-7×10) / 2π×0.01 = 8×10^-17N.
(2) A galvanometer having resistance 3Ω is converted into an ammeter by connecting a 3Ω shunt resistance across it. In order to double the range of this ammeter, the additional shunt resistance to be connected across it is
(a) 3Ω (b) 2.5Ω (c) 2Ω (d) 1.5Ω (e) 6Ω
With usual notations, Ig G = (I-Ig)S. Since G = S =3Ω, Ig = I/2. If S′ is the additional shunt required, the effective shunt for doubling the range is SS′/(S+S′) . Therefore, when the range is doubled, IgG = (2I-Ig) SS′/(S+S′) . Substituting G=S=3 and Ig= I/2 , 3S'/(3+S') =1. Therefore,S'=1.5Ω.
Instead of following the above mathematical steps, you can easily arrive at the answer as follows:
The ammeter with the original shunt has an effective resistance of 1.5Ω. If you want to pass twice the maximum allowed current into the ammeter, half of this doubled current has to be diverted through another path. So you have to provide a shunt path of resistance 1.5Ω.
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