Physics-
General
Easy

Question

A wire shown in figure carries a current of 40 A. If r equals 3.14 blank c m, the magnetic field at point P will be

  1. 1.6 cross times 10 to the power of negative 3 end exponent T    
  2. 3.2 cross times 10 to the power of negative 2 end exponent T    
  3. 4.8 cross times 10 to the power of negative 3 end exponent T    
  4. 6.0 cross times 10 to the power of negative 4 end exponent T    

The correct answer is: 6.0 cross times 10 to the power of negative 4 end exponent T


    Let the given circular A B C part of wire subtends an angle theta at its centre. Then, magnetic field due to this circular part is

    B to the power of ´ end exponent equals B subscript c end subscript cross times fraction numerator theta over denominator 2 pi end fraction equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction cross times fraction numerator 2 pi i over denominator e end fraction cross times fraction numerator theta over denominator 2 pi end fraction
    rightwards double arrow blank B to the power of ´ end exponent equals blank fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction blank. fraction numerator i over denominator r end fraction theta
    Given, i equals 40 blank A comma blank r equals 3.14 blank c m equals 3.14 cross times 10 to the power of negative 2 end exponent blank m
    theta equals 360 degree minus 90 degree equals 270 degree equals fraction numerator 3 pi over denominator 2 end fraction blank r a d.
    therefore B to the power of ´ end exponent equals fraction numerator 10 to the power of negative 7 end exponent cross times 40 over denominator 3.14 cross times 10 to the power of negative 2 end exponent end fraction cross times fraction numerator 3 pi over denominator 2 end fraction
    B to the power of ´ end exponent equals 6 cross times 10 to the power of negative 4 end exponent T

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