Question

The sphere is nonconducting, has a mass of 72.0 g, and a radius 19.0 cm. A flat, compact coil of wire with five turns is wrapped tightly around it, with each turn concentric with the sphere. The sphere is placed on the incline so that the coil is parallel to the plane. You establish a uniform magnetic field of 0.315 T vertically upward in the region of the sphere.

a. What current in the coil will enable the sphere to rest in equilibrium on the inclined plane?
b. Show that the result does not depend on the value of θ.

Answer 1

a) 0.751 A

b) From the derivation below, it is glaring that the current (I) is independent of θ

Step-by-step explanation:

Given data :

mass of the sphere (m) = 72.0 g = 0.072 kg

number of turns of the wire (N) = 5

radius of the sphere (r) = 19.0 cm = 0.19 m

uniform magnetic field (B) = 0.315

From the given question, the sphere is in translational equilibrium; therefore
`F_s-Mg sin \theta=0`

`F_s=Mg sin \theta` ----------Equation (1)

At the center of the sphere, Lets look at the consideration of the torque; two things is happening;

1) the magnetic field produces a clockwise torque of magnitude μBsinθ

2) the frictional force produces a counterclockwise torque of magnitude
`F_sr`

∴

`F_sr` - μBsinθ = 0 ---------- Equation (2)

Substituting
`F_s = (Mgsin \theta)` from equation (1) to equation (2); we have:

mgsinθr - μBsinθ = 0

Dividing both sides by sinθ; we have:

mgr - μB = 0

mgr = μB

μB = mgr ------------------- Equation (3)

Given that:

μ = NIπr²

Substituting μ = NIπr² into Equation (3); we have:

NIπr²B = mgr

Dividing both sides by r; we have:

NIπrB = mg

Making (I) =
`(mg)/(N\pi rB)`

Substituting our given data into equation above; we have:

(I) =
`((0.072kg)(9.8))/((5*\pi )(0.19 m)(0.315))`

(I) = 0.751 A in counterclockwise direction.

b)

Show that the result does not depend on the value of θ.

From equation (2); we have:

`F_sr` - μBsinθ = 0

We also know that;

`F_s = (Mgsin \theta)` ;

so replacing
`F_s = (Mgsin \theta)` into above equation (2); we have:

`Mgsin \theta r` - μBsinθ = 0

`Mgsin \theta r` = μBsinθ

Dividing both sides by sinθ; we have:

Mgr = μB

However; μ = NIπr²

Substituting it into the above derivation; we have:

Mgr = NIπr²B

Dividing both sides by r; we have:

Mg = NIπrB

Making (I) the subject of the formula; we have:

I =
`(mg)/(N\pi rB)`

∴ we can conclude that the current is independent on the value of θ.