Question

The wire shown in figure carries a current of 40 A. If r = 3.14 cm the magnetic field at point p will be:

(A) $1.6 \times {10^{ - 3}}T$
(B) $3.2 \times {10^{ - 3}}T$
(C) $6 \times {10^{ - 4}}T$
(D) $4.8 \times {10^{ - 3}}T$

Answer 1

Hint: To answer this question we have to begin with the formula for finding the magnetic field at a circular arc. Then we have to evaluate the expression and put the values in the expression from the question. This will give us the required answer to the question.

Complete step by step answer:
We should know that the magnetic field occurring due to the circular arc will be given by:
$B = \dfrac{{{\mu _0}I}}{{4\pi r}}\theta $
Here I represent the current, r is the radius of the circular arc, ${\mu _0}$is the permeability of the medium in the vacuum and $\theta $ is the angle which is subtending from that of the centre.
Now we have to evaluate the above formula. So here it is:
$B = \dfrac{{{\mu _0}I \times 3\pi }}{{4\pi r \times 2}} = \dfrac{{3{\mu _0}I}}{{8r}}$
Now we have to put the values in the above expression to get the following:
$
  B = \dfrac{{3 \times 4\pi \times {{10}^{ - 7}} \times 40}}{{8 \times 3.14 \times {{10}^{ - 2}}}} \\
   \Rightarrow B = 6 \times {10^{ - 4}}T \\
 $
Hence we can say that the magnetic field at the point p is given as $6 \times {10^{ - 4}}T$.

So the correct option is option C.

Note: We have come across the term magnetic field in this answer. For the better understanding we need to familiarize ourselves with the definition. So by the definition magnetic field is defined as the vector field which is best described to have the magnetic influence on an electric charge of various moving charges or magnetized materials.
We should know that a charge which is moving in a magnetic field will always experience a force which is in perpendicular to the velocity of itself and to that of the magnetic field.