(A Level Phy) Deriving gravitational field strength, electric field strength and magnetic flux density due to 3 rings

Not only the ring can inspire many memorable movies such as THE LORD OF THE RINGS, it can also be a source of challenging problems in gravitational field, electric field and magnetic field. Let's derive three expressions for the gravitational field strength g, electric field strength E and magnetic flux density B due to a ring of mass M, a ring of charge Q and a current-carrying loop of length L respectively. These three problems are not related to one another. But I put them side by side for your comparison.

Laws to use:

Newton's Law of Gravitation: F = Gmm/r²

Coulomb's Law: F = Q₁Q₂/4πε_or²

Biot-Savart Law: B = µ_oI L sin θ/4πr²

Gravitational field strength g due to a ring of mass M	Electric field strength E due to a ring of charge Q	Magnetic flux density B due to a current-carrying loop
A unit mass is at P which is along an x-axis that passes through the centre of a ring. The ring of mass M and radius a exerts a gravitational field g on the unit mass. Take a small mass ∆M on the ring. Let the distance between m and ∆M be s. From Newton's Law of Gravitation, the gravitational field strength ∆g acting by ∆M = -G∆M/s2 where G is the gravitational constant We can resolve ∆g into ∆gx and ∆gy. If the angle between ∆g and x axis is α, then ∆gx = ∆g cos α. But cos α = x/s. So, ∆gx = ∆g cos α = -G∆M cos α /s2 = -G∆M x/s3 The ring is symmetrical. For every ∆M on one side, there is an opposite ∆M such that the two vertical components of ∆g cancel each other. The resultant ∆g exerted by the ring is therefore along the x-axis. Sum up all the individual ∆g exerted by the individual ∆M, the gravitational force on the unit mass by the ring M is: g = -GMx/s3	A unit positive charge is at P which is along an x-axis that passes through the centre of a ring. The ring of charge +Q and radius a exerts an electric field E on the unit charge. Take a small charge ∆Q on the ring. Let the distance between P and ∆Q be s. From Coulomb's Law, the small electric field ∆E acting by ∆Q = k∆Q/s2 where k = 1/4πεo We can resolve ∆E into ∆Ex and ∆Ey. If the angle between ∆E and x-axis is α, then ∆Ex = ∆E cos α. But cos α = x/s. So, ∆Ex = ∆E cos α = k∆Q cos α /s2 = k∆Q x /s3 The ring is symmetrical. For every ∆Q on one side, there is an opposite ∆Q such that the two vertical components of ∆E cancel each other. The resultant ∆E exerted by the ring is therefore along the x-axis. Sum up all the individual ∆E exerted by the individual ∆Q, the electric field strength on the unit charge by the ring Q is: E = kQx/s3	P is along an x-axis that passes through the centre of a circular loop of radius a. The loop carries a current I flowing in an anti-clockwise direction from the point of view of P. Take a small section ∆L at the top of the ring. Here, I points out of the page. Using the right-hand grip rule with your thumb pointing out of the page, the magnetic field ∆B at P is diagonally upward at right angle to s. By Biot-Savart Law, ∆B = µoI ∆L sin θ/4πs2 where θ is the angle between ∆L and s Since θ = 90º, sin θ = 1. So, ∆B = µoI ∆L/4πs2 Consider only the field along the x-axis. So we resolve ∆B into ∆Bx and ∆By. If the angle between ∆B and the x-axis is α, then ∆Bx = µoI ∆L cos α /4πs2 The ring is symmetrical. For every ∆L on one side, there is an opposite ∆L such that the two vertical components of ∆B cancel each other. The resultant ∆B exerted by the ring is therefore along the x-axis. Sum up all the individual ∆B by the individual ∆L, the magnetic field at P is: B = µoI L cos α /4πs2 where L is the circumference of the loop

Try these questions before you check with the answers below. You can directly use the results derived above.

1. Define gravitational field strength.
   
2. Show that the gravitational force acting on a mass m by a ring of mass M is F = -GMmx/(x² + a²)^3/2 where a = radius of the ring, x = distance between m and the centre of the ring.
   
3. Define electric field strength.
   
4. Show that the electric force acting on a positive charge q by a ring of charge Q is qaσx/[2ε_o(x² + a²)^3/2] where σ = charge density = Q/2πa and a = radius of the ring, x = distance between q and the centre of the ring.
   
5. Define magnetic flux density.
   
6. Show that the magnetic flux density at a point P along the x-axis due to a current-carrying loop is µ_oI a² /2(x² + a²)^3/2 where a = radius of the loop, I is the current in the loop, x is the distance between P and the centre of the loop.

Answers: 1. The gravitational field strength at a point in a field is the force per unit mass acting at that point. 2. We have derived g = -GMx/s³ ==> F = mg = -GmMx/s³. Let's express s in terms of a and x. By Pythagoras Theorem, s² = x² + a² ==> s = (x² + a²)^½. Therefore, ∆F_x = -Gm∆M x/(x² + a²)^3/2. F = -GmMx/(x² + a²)^3/2. 3. Electric field strength is the force per unit positive charge acting at a point in the field. 4. We have derived E = kQx/s³. Let's express s in terms of a and x. By Pythagoras Theorem, s² = x² + a² ==> s = (x² + a²)^½. Therefore, ∆F_x = kq∆Q x/(x² + a²)^3/2. F = kqQx/(x² + a²)^3/2. If Q = 2πaσ where σ = charge density, then F = q(2πaσ)x/[4πε_o(x² + a²)^3/2] = qaσx/[2ε_o(x² + a²)^3/2]. 5. Magnetic flux density is the amount of flux in a unit area perpendicular to the direction of magnetic flow. 6. L = 2πa, cos α = a/s, and by Pythagoras Theorem s = √(a² + x²). We have derived B = µ_oI L cos α /4πs² = B = µ_oI 2πa (a/s) /4πs² = µ_oI a² /2s³ = µ_oI a² /2(x² + a²)^3/2