Physics-
General
Easy

Question

A particle of specific charge alpha (charge per unit mass) is released at time t=0 from origin with an initial velocity of stack v with rightwards arrow on top equals v subscript 0 end subscript stack i with ˆ on top in a uniform magnetic field stack B with rightwards arrow on top equals negative B subscript 0 end subscript stack k with ˆ on top. Find position of particle at any time t.

  1. fraction numerator v subscript 0 end subscript over denominator B subscript 0 end subscript alpha end fraction open square brackets S i n invisible function application open parentheses B subscript 0 end subscript alpha t close parentheses stack i with hat on top minus open square brackets 1 plus C o s invisible function application open parentheses B subscript 0 end subscript alpha t close parentheses stack j with hat on top close square brackets close square brackets    
  2. fraction numerator v subscript 0 end subscript over denominator B subscript 0 end subscript alpha end fraction open square brackets S i n invisible function application open parentheses B subscript 0 end subscript alpha t close parentheses stack i with hat on top plus open curly brackets 1 minus C o s invisible function application open parentheses B subscript 0 end subscript alpha t close parentheses stack j with hat on top close curly brackets close square brackets    
  3. fraction numerator v subscript 0 end subscript over denominator B subscript 0 end subscript alpha end fraction open square brackets C o s invisible function application B subscript 0 end subscript alpha t plus open curly brackets 1 minus S i n invisible function application open parentheses B subscript 0 end subscript alpha t close parentheses stack j with hat on top close curly brackets close square brackets    
  4. fraction numerator v subscript 0 end subscript over denominator B subscript 0 end subscript alpha end fraction left square bracket S i n invisible function application alpha t stack i with hat on top minus left curly bracket 1 plus C o s invisible function application left parenthesis alpha t right parenthesis stack j with hat on top right curly bracket right square bracket    

The correct answer is: fraction numerator v subscript 0 end subscript over denominator B subscript 0 end subscript alpha end fraction open square brackets S i n invisible function application open parentheses B subscript 0 end subscript alpha t close parentheses stack i with hat on top plus open curly brackets 1 minus C o s invisible function application open parentheses B subscript 0 end subscript alpha t close parentheses stack j with hat on top close curly brackets close square brackets

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