Question

# An electron gun G emits electrons of energy 2 keV travelling in the positive x-direction. The electrons are required to hit the spot S where GS=0.1m, and the line GS makes an angle of 60o with the x axis as shown in fig. A uniform magnetic field →B parallel to GS exists in the region outside the electron gun. If the minimum value of B needed to make the electrons hit S is Bmin=4.73×10−XT. find X?

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Solution

## Kinetic energy of electron, K=12mv2=2keVSpeed of electron, v=√2Km=√2×2×1.6×10−169.1×10−31ms−1=2.65×107ms−1Since, the velocity (→v) of the electron makes an angle of θ=60o with the magnetic field →B, the path will be a helix.So, the particle will hit S if Gs=np; Here, n=1,2,3,....P= pitch of helix =2πmqBvcosθ⇒GS=n2πmvcosθqB⇒B=n2πmvcosθq(GS)But for B to be minimum, n=1Bmin=2πmvcosθq(GS)Substituting the values, we have:Bmin=(2π)(9.1×10−31)(2.67×107)(12)(1.6×10−19)(0.1)Bmin=4.73×10−3T

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