A current I is flowing through the loop as shown in the . . .

Question : A current I is flowing through the loop as shown in the figure (MA=R, MB=2R). The magnetic field at the centre of the loop is µ₀I/R times:

a) 5/16, into the plane of the paper. 

b) 5/16, out of the plane of the paper. 

c) 7/16, out of the plane of the paper. 

d) 7/16, into the plane of the paper. 

Doubt by Trishna

Solution : 

We know, magnetic field due to a circular current carrying loop at the centre is given by 
B = µ₀I/2r

And the magnetic field due to an circular current carrying arc (θ) is given by 

B = (θ/360)×[µ₀I/2r]

Now
B₁ = (270/360)×[µ₀I/2R]
B₁ = (3/4)×[µ₀I/2R] — (1)
By using, Right Hand Thumb rule, the direction of magnetic field will be into the plane of the paper. 

B₂=(90/360)×[µ₀I/2(2R)]
B₂=(1/4)×[µ₀I/4R] — (2)

By using, Right Hand Thumb rule, the direction of magnetic field will be into the plane of the paper. 

The magnetic field due to straight wires AB and DC will be zero as θ is either 0° or 180° and by using Biot-Savarts law magnetic field will be zero. 

Total Magnetic field at the centre of the loop will be given by adding equation (1) and (2)

B₁+B₂=(3/4)×[µ₀I/2R]+(1/4)×[µ₀I/4R]

B = µ₀I/R[3/8+1/16]
B = µ₀I/R[6/16+1/16]

B = µ₀I/R[7/16]
B = (7/16)×(µ₀I/R)

Direction of Total Magnetic Field will be into the plane of the paper. 

Hence, d) 7/16, into the plane of the paper, would be the correct option.