Resultant field at the center of two semicircular current arcs

[tanaygupta2000]

Homework Statement

A wire carrying a current 2A is bent into the shape of two semi-circles, one with a radius 1m and the other with radius 2m. The two semi-circles are connected at their ends that is along the two radii. The magnetic field at the center of the semi-circles is?
(a) πe-7 T
(b) Cannot be determined from the information provided above
(c) 16e-7 T
(d) 2e-7 T

Relevant Equations

Field at the center of a current loop, B = μ0 i /2r

So the magnetic field induced at the center of a current-carrying loop is given by:
B = μ0 i /2r
where r is the radius of the loop

In the case of a semi-circular loop, this becomes
B = μ0 i /4r

In the question, i = 2A, r1 = 1m and r2 = 2m
So, field induced at the center of first semicircular loop is given by,
B1 = μ0 × 2 /4 × 1 = μ0/2
and
B2 = μ0 × 2 /4 × 2 = μ0/4

Now I'm not getting what the question is asking. If it is asking the resultant field in the case of two semicircular wires connected in the same plane, then we will have,
B = B1 + B2 = 3μ0/4 = 9.42e-1 T

and if it is asking the resultant field in the case of two semicircular wires connected with their planes perpendicular to each other, then we will have
B = sqrt(B1^2 + B2^2) = 14.04e-7 T

But these both options are not available.
Kindly help!
(The correct option is 16e-7 T, i.e., option-(c))

 

[Delta2]

Can you provide an image because from the description i can't understand completely what is the exact geometry of the setup.

 

[ProxyMelon]

I would say that the two semi-circles can be in two shapes (they are likely on the same plane), either forming a C (in this case the point in which you want to evaluate B is outside of the loop) or forming an O (in this case it is inside the loop).

How whould things chance in the two setup (also how would things change considering the current flowing clockwise or anti-clockwise, so 4 setups in total)?

 

[haruspex]

You can rule out the orthogonal arrangement since the connecting wires would not be radial.
As @ProxyMelon points out, this still leaves two possibilities: that the two semicircles face each other, so the fields add, or one nests inside the other, so the fields oppose.
If there is a diagram and it shows the nested arrangement, that matches one answer. If there is no diagram then it matches a different one.

 

[tanaygupta2000]

This is a rough sketch I've drwin according to my partial understanding.
I'm not getting the meaning of "The two semi-circles are connected at their ends that is along the two radii. The magnetic field at the center of the semi-circles is?" line.

 

[tanaygupta2000]

The correct option is 16e-7 T, i.e., option-(c)

 

[haruspex]

This is a rough sketch I've drwin according to my partial understanding.
I'm not getting the meaning of "The two semi-circles are connected at their ends that is along the two radii. The magnetic field at the center of the semi-circles is?" line.

That's not how I read it. Your diagrams have arcs, but they are not semicircles.

Draw an arc from 12 o'clock to 6 o'clock. At a different radius with the same centre, draw another arc between those times. Then connect the ends of the arcs with wires radially along the line from 6 to 12.
But there are two possibilities. You can have both arcs on the same same side, or on opposite sides forming a letter C. One of these leads to one of the options.

Edit: whoops, I wrote that backwards. I meant "You can have both arcs on the same same side, forming a letter C, or on opposite sides."

 

[tanaygupta2000]

That's not how I read it. Your diagrams have arcs, but they are not semicircles.

Draw an arc from 12 o'clock to 6 o'clock. At a different radius with the same centre, draw another arc between those times. Then connect the ends of the arcs with wires radially along the line from 6 to 12.
But there are two possibilities. You can have both arcs on the same same side, or on opposite sides forming a letter C. One of these leads to one of the options.

Yes sir, for arcs on the opposite side, net field due to two semicircles subtract, giving B1-B2 = μ0/4 = πe-7 T (option a)

 

[haruspex]

Yes sir, for arcs on the opposite side, net field due to two semicircles subtract, giving B1-B2 = μ0/4 = πe-7 T (option a)

Despite my blunder (see edited post #7) you seem to have figured out what I meant.