COMEDK Exam 2018 Solved Paper

The given current carrying conductor is shown belowMagnetic field at the centre due to circular wire,
B1=‌

µ0I

2r

×‌

270

360

=‌

µ0I

2r

×‌

3

4

B1=‌

3µ0I

8r

[in downward to the plane of paper]
Magnetic field at centre O due current carrying wire AB will be zero because point O lies along AB
∴‌‌B2=0
Magnetic field at O due to current carrying wire CD B3=‌

µ0I

4πr

(in downward to the plane of paper)
∴ Net magnetic field at the centre,
B‌net ‌=B1+B2+B3=‌

3µ0I

8r

+0+‌

µ0I

4πr

=‌

µ0I

r

[‌

3

8

+‌

1

4π

]=‌

µ0I

4πr

[‌

3π

2

+1]