Question

A current i is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure.The magnetic field at the centre of the loop is, μ0iR times[ MA = R , MB = 2R, ∠DMA = 90°]

• A. 516 , but out of the plane of the paper
• B. 516, but into the plane of the paper
• C. 716, but out of the plane of the paper
• D. 716, but into the plane of the paper

Answer 1

Answer: (D)

716, but into the plane of the paper

Answer 2

Explanation:
Given:
MA = R, MB = 2R, ∠ DMA = 90°We have to find the magnetic field at the center of the loop.The magnitude of magnetic field due to an arc of radius R and angle θ, carrying current at the center of the loop isB = πμ0i4πRθwhere, θ is the angle subtended by the arc.Given figure is redrawn below:The magnetic field due to curve DA isBDA=πR(3π2)
Similarly, magnetic field due to the arc BC isBBC=π2R(π2)[ given MB=2R]
Also, the magnetic field due to the curve DC and AB isBDC=BAB=[since the angle between the current element and the position vector is zero, i.e. dI⇀ × r⇀ =0
Now, the resultant magnetic field at the centre M isBnet =BDA+BAb+BBC+BCD
⇒Bnet=πR(3π2)++8πR(π2)+
⇒Bnet=R[32+1]⇒Bnet=R×[6+1]⇒Bnet=716R
Using the right-hand thumb rule for the magnetic field, the direction of the magnetic field is into the plane of the paper, as the direction of the current is clockwise.
Hence, the correct option is (D).