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A loop ABCDA, carrying current I = 12 A, is placed in a plane, consists of two semi-circular segments of radius $R_1 = 6\pi$ m and $R_2 = 4\pi$ m. The magnitude of the resultant magnetic field at center O is $k\times10^{-7}$ T. The value of k is __________ 

(Given $\mu_0 = 4\pi \times 10^{-7}$ Tm $A^{-1}$)

To solve this problem, we need to find the magnetic field at the center O due to the current-carrying loop ABCDA, which includes two semi-circular segments.

The magnetic field at the center due to a current-carrying wire in a semicircular arc is given by:

B = (μ₀I)/(4R)

Step 1: Calculate the magnetic field due to the larger semicircle (radius R1)
The magnetic field for R1 = 6π m is:

B1 = (4π × 10-7 Tm A-1 × 12 A) / (4 × 6π m)

= (48π × 10-7) / (24π) T

= 2 × 10-7 T

Step 2: Calculate the magnetic field due to the smaller semicircle (radius R2)
The magnetic field for R2 = 4π m is:

B2 = (4π × 10-7 Tm A-1 × 12 A) / (4 × 4π m)

= (48π × 10-7) / (16π) T

= 3 × 10-7 T

Step 3: Determine the net magnetic field at the center O
The directions of the magnetic fields induced by the two segments are opposite because they are on opposite sides of O. Therefore, the net field is:

Bnet = B1 - B2
= 2 × 10-7 T - 3 × 10-7 T
= -1 × 10-7 T

The negative indicates direction, but since we are asked for magnitude, we consider positive:
Bnet = 1 × 10-7 T

Step 4: Determine the value of k
Since Bnet = k × 10-7 T, comparing gives k = 1.

The calculated value of k = 1, which fits within the given range of 1,1.

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Important Questions from Electricity and Magnetism

  1. Two blocks of masses $m$ and $M$, ($M > m$), are placed on a frictionless table as shown in figure. A massless spring with spring constant $k$ is attached with the lower block. If the system is slightly displaced and released, then

     

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