"Seven Deadly Sins:

Wealth without work

Pleasure without conscience

Science without humanity

Knowledge without character

Politics without principle

Commerce without morality

Worship without sacrifice."

– Mahatma Gandhi

The following multiple choice question included in the All India Engineering/Architecture Entrance Examination (AIEEE) 2011 will be interesting and useful to you:

A current I flows in an infinitely long wire with cross section in the form of a semi-circular ring of radius R. The magnitude of the magnetic induction along its axis is

(1) μ_{0 }I/4πR

(2) μ_{0 }I/π^{2}R

(3) μ_{0 }I/2π^{2}R

It would have been better if the magnitude of the magnetic induction *at the axis* was asked for since the words *along the axis* usually means *directed along the axis*. [The component of field *directed along the axis* is zero].

Forget about it. The question setter requires you to calculate the magnetic flux density at points on the axis (of the semicircular ring shaped cross section of the infinitely long wire). In the adjoining figure we have shown the cross section (of the given wire) lying in the XY plane. The length of the wire is along the Z-axis and the current in the wire is supposed to flow along the negative Z-direction. The broad infinitely long wire can be imagined to be made of a large number of infinitely long straight wire strips, each of small width dℓ.

With reference to the figure, we have dℓ = Rdθ.

The magnetic flux density due to the above strip is shown as d**B**_{1} in the figure. It has an X-component dB** _{1 }**sinθ and a Y-component dB

**cosθ. When we consider a similar strip of the same with dℓ**

_{1}*located*

*symmetrically with respect to the Y-axis*, we obtain a contribution dB

_{2 }to the flux density. The flux density d

**B**

_{2}has the

*same*magnitude as d

**B**

_{1}. It has

_{ }X-component dB

_{2 }sinθ and Y-component dB

_{2}cosθ. The X-components of d

**B**

_{1}and d

**B**

_{1}are of the same magnitude and direction and they add up. But the Y-components of d

**B**

_{1}and d

**B**

_{1}are in

*opposite*directions and have the

*same*magnitude. Therefore they get canceled. The entire conductor therefore produces a resultant magnetic field along the negative X-direction.

The wire strip of width dℓ cn be imagined to be an ordinary thin straight infinitely long wire carrying current Idℓ/πR since the total current I flows through the semicircular cross section of perimeter πR. Putting dB_{1} = dB_{2} = dB we have

dB = μ_{0} (Idℓ/πR)/2πR = μ_{0} (IRdθ/πR)/2πR = μ_{0}Idθ/2π^{2}R

The X-component of the above field is (μ_{0}I/2π^{2}R) sinθ dθ

The field due to the entire conductor is B = _{0}∫^{π }[(μ_{0}I/2π^{2}R) sinθ]dθ

Or, B = μ_{0}I/π^{2}R since _{0}∫^{π }sinθ dθ = 2