Answer to Question #112923 in Electricity and Magnetism for Walid

Question

Find the magnetic field at the point O, if the current is = 100 А and =20 cm.
2. The thin wire carrying the current =10 A is bent into the rectangle with sides 6 cm and =10.5 cm. Find the magnetic field at the point of diagonals intersection.
3. A horizontal wire carries a current of 48 A towards the east. A second wire with mass 0.05 kg runs parallel to the first, but lies 15 cm below it. This second wire is held in suspension by the magnetic field of the first wire above it. If each wire has a length of half a meter, what is the magnitude and direction of the current in the lower wire?
4. An electron having 880 eV of energy moves at right angles to a uniform magnetic field of flux density 2.5·10-3 T. Show that the path of the electron is a circle and find its radius.

Answer 1

1. In a circular loop, the field at the center will be

$B=\frac{\mu_0I}{2R}=\frac{4\pi\cdot10^{-7}\cdot100}{2\cdot0.2}=3.14\cdot10^{-4}\text{ T}.$

2. In a rectangular configuration, each loop creates a field at the center

$B_r=\frac{\mu_0I}{2ab}\sqrt{a^2+b^2}.$

Two such rectangles produce

$B=2B_r=\frac{\mu_0I}{ab}\sqrt{a^2+b^2}=\\ =\frac{4\pi\cdot10^{-7}\cdot10}{0.06\cdot0.105}\sqrt{0.06^2+0.105^2}=2.96\cdot10^{-4}\text{ T}.$

3. The field of the first top wire at a distance of 15 cm is

$B_t=\frac{\mu_0I_t}{2\pi R}.$

The force that is required to compensate the force of gravity of the second wire,on the one hand, is

$F=mg,$

and, on the other:

$F=I_bB_tL,\\ I_b=\frac{F}{B_tL}=\frac{mg}{B_tL}=\frac{2\pi mgR}{L\mu_0I_t} =\\ \space\\ =\frac{2\pi 0.05\cdot9.8\cdot0.15}{0.5\cdot4\pi\cdot10^{-7}\cdot48}=15313\text{ A}.$

We know that currents flowing in the same direction attract, therefore, the current in the lower wire must flow towards the east.

4. Find the speed of the electron:

$E_K=\frac{1}{2}mv^2, \\ v=\sqrt{\frac{2E_K}{m}}.$

The electron experiences Lorentz's force, which is equal to the centripetal force:

$F=evB=m\frac{v^2}{R},\\ \space\\ R=\frac{mv}{eB}=\frac{\sqrt{2mE_K}}{eB}=\\ \space\\=\frac{\sqrt{2(9.11\cdot10^{-31})(880\cdot1.6\cdot10^{-19})}}{(1.6\cdot10^{-19})(2.5·10^{-3})}=0.04\text{ m}.$