Question

The two wire loops ABCD formed by joining two semicircular wire radii $\sf{{R}_{1}}$ and $\sf{{R}_{2}}$ carries a current I, as shown in the above figure. The magnetic induction at the centre O is:
A) $\sf{\dfrac{{μ}_{0}I({R}_{2}-{R}_{1})}{4{R}_{1}{R}_{2}}}$ out of the page
B) $\sf{\dfrac{{μ}_{0}I({R}_{2}-{R}_{1})}{4({R}_{1}{R}_{2})}}$ into the page
c) $\sf{\dfrac{{μ}_{0}I{R}_{1}{R}_{2}}{4{R}_{2}-{R}_{1}}}$ out of the page
d) Zero
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Answer 1

Answer:

option: a)

Explanation:

The resultant magnetic field at O will be the sum of the magnetic fields due to the current in the two semicircles, and we can use the expression for the magnetic field at the center of a current loop to find B.The resultant magnetic field at point O is given as B = B1+B2

The magnetic field at the center of a current loop is B = 2πRμ0I, where R is the radius of the loop.

The magnetic field at the center of half a current loop

B =212Rμ0I=4Rμ0I.

Therefore,

B1 = 4R1μ0I and B2 = - 4R2μ0I.

So, B = B1+B2 = 4R1μ0I - 4R2μ0I.

That is, 4μ0I(R11−R21)=μ0I(4R1.R2R2−R1).

Hence, the magnetic induction at the center is μ0I(4R1.R2R2−R1) out of the page.

Answer 2

Answer:

Correct option is A)

Answer is A.

Explanation:

hope it helps you