Question

A long hairpin is formed by bending a very long wire, as shown. If the wire carries a current of 15 A, what is the magnitude and direction of the magnetic field at the point a? Take R = 15 mm. Take the sign of B to be positive if the direction is out of the paper and negative if B is into the paper. You need to enter the number with its sign and space and the unit.

What is the magnitude and direction of the magnetic field at the midpoint b? The distance between a and b are much larger than R (each straight section is "infinite"). Take the sign of B to be positive if the direction is out of the paper and negative if B is into the paper.

Answer 1

Given is:-

The current is the wire is  

=15A

Radius of the curve   or $R = 15 \times 10^{-3}m$

Now,

The magnetic field due to the semi straight wire is given by

$B = \frac{\mu_o I}{4 \pi R}$

and the magnetic field at the centerr of the circular arc is given by  $B = \frac{\mu_o I}{4 R}$

therefore the magnetic field at point a is given by

$B = 2 \frac{\mu_o I}{4 \pi R}+\frac{\mu_o I}{4 R}$

or

$B = \frac{\mu_o I}{2R}(\frac{1}{\pi} + \frac{1}{2})$

by plugging the values we get

$B = \frac{(4 \pi \times 10^{-7}) (15A)}{2(15 \times 10^{-3}m)}(\frac{1}{\pi} + \frac{1}{2})$

which gives us

$\boxed{B = 0.49817mT}$

And the magnetic field at the point b is given by

$B =2 (\frac{\mu_o I }{2 \pi R})$

or

$B =2 (\frac{(4 \pi \times 10^{-7}) (15A) }{2 \pi (15 \times 10^{-3}m)})$

which gives us

$\boxed{B =0.4mT}$