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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

$1.6 \times 10^{- 3} T$
$3.2 \times 10^{- 2} T$
$4.8 \times 10^{- 3} T$
$6.0 \times 10^{- 4} T$

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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