Physics-
General
Easy

Question

A proton accelerated by a potential difference 500 blank K V moves though a transverse magnetic field of 0.51 blank T as shown in figure. The angle theta through which the proton deviates from the initial direction of its motion is

  1. 15 degree    
  2. 30 degree    
  3. 45 degree    
  4. 60 degree    

The correct answer is: 30 degree


    According to the following figure, sin invisible function application theta equals fraction numerator d over denominator r end fraction

    Also r equals fraction numerator square root of 2 m k end root over denominator q B end fraction equals fraction numerator 1 over denominator B end fraction square root of fraction numerator 2 m V over denominator q end fraction end root
    therefore sin invisible function application theta equals B d square root of fraction numerator q over denominator 2 m V end fraction end root
    equals 0.51 cross times 0.1 square root of fraction numerator 1.6 cross times 10 to the power of negative 19 end exponent over denominator 2 cross times 1.67 cross times 10 to the power of negative 27 end exponent cross times 500 cross times 10 to the power of 3 end exponent end fraction end root
    equals fraction numerator 1 over denominator 2 end fraction rightwards double arrow theta equals 30

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