Question

A 3.0 m long straight horizontal wire carrying a current of 4.7 A along the +x-direction is in a region of uniform magnetic field, B. If the mass of this wire is 0.4 kg, find the magnitude and direction of the magnetic field necessary to have this wire "float" in the Earth's gravitational field. (Assume the force due

Answer 1

Answer:

B=18.15T

Explanation:

The force due to the magnetic field must equal the gravitational force

Fm=Fg

We assume that the direction of the magnetic field is perpendicular to the current in the wire. By using the expression for the magnetic force in a wire, we have

$F=iLBsin\theta=iLB$

where L is the length of the wire, and i is the current.

For the gravitational force we have:

$F=Mg=(0.4kg)(9.8\frac{m}{s^2})=3.92N$

Hence, by doing Fm=Fg

$F_{m}=F_{g}\\\\iLB=256N\\\\B=\frac{256N}{(4.7A)(3.00m)}=18.15T$

Hope this helps!!

Answer 2

Answer:

B = 0.2780 T

The direction is along positive + z-direction

Explanation:

Given that :

A 3.0 m long straight horizontal wire carrying a current of 4.7 A

i.e L = 3.0 m

I = 4.7 A

Which is along x - axis direction in a uniform magnetic (B) field

mass of the wire = 0.4 kg

So;

$F_B = F_g$

$B *I*L(sin 90) = mg$

$B = \frac{mg}{I*L*(sin90)}$

$B = \frac{0.4*9.8}{4.7*3.0}$

$B = \frac{3.92}{14.10}$

B = 0.2780 T

And:

The direction is along positive + z-direction