Magnetic fieldn at the centre of a circular coil of radius R carrying current 'I' is B `alpha (1)/(R )` and its direction is given by right - hand thumb rule. Magnetic field at the centre of a circular are subtending an angle 0 (in degree) is `Ba (1)/(R ) ((theta)/(360^(@)))` and its direction can be found using right hand rule.

Consider two circular coils made of uniform conductors as shown in figure 3 and 4. In figure 3, points C and D are diametrically opposite to each other and in figure 4, `angle PO = Q = 120^(@)` . Then magnetic fields __________ .

Options

A. At both `O_(1) and O_(2)` are zero.
B. At both `O_(1) and O_(2)` are non-zero
C. At zero at `O_(1)` but non-zero at `O_(2)`
D. Is non-zero at `O_(1)` but zero at `O_(2)`

Answer

A

Explanation

In figure 3,
Magnetic field `Balpha(1)/(R)[(theta)/(360^(@))]`
`i_(i)=i[(piR_(2))/(2piR)]=(i)/(4),i_(2)=i[((3piR)/(2))/((2)/(2piR))]=(3)/(4)i`
Since, we have been given that magnetic field `B_(2)=K(i_(2))/(2)[(270^(@))/(360^(@))]["into to place"].......(i)`
`B_(2)=K(i_(2))/(R)[(90^(@))/(360^(@))]["out of plane"]......(ii)`
where, `2i_(i) = 3i_(i)`
On dividing eqns (i) by (ii), we get
Now,
`(B_(1))/(B_(2))=(ki_(2)R)/(R.ki_(2))[(270^(@))/(360^(@))]`
`rArr (B_(1))/(B_(2)) = 1`
Therefore, net field is zero at `O_(1)` is 0
In the same way, from figure, `4, i_(4) = 2i`,
and `B_(3)(ki_(3))/(R)[(240^(@))/(360^(@))]["into the plane"]=......(iii)`
`B_(4)=(ki_(4))/(R)[(120^(@))/(360^(@))]["out of the plane"]........(iv)`
On dividing eqns (iii) by (iv) we get
Now, `(B_(3))/(B_(4)) = 1`
Therefore net field at `O_(2)` is zero.