Question Details

In the circuit shown below, a current 3 I enters at A. The semicircular parts ABC and ADChaveequal radii r but resistances 2R and R respectively. The magnetic field at the center of the circular loop ABCD is ________.


Options

A

µ0I/4r out of the plane

B

µ0I/4r into the plane

C

µ03I/4r out of the plane

D

µ03I/4r into of the plane

Show Answer

Correct Answer :

Option D

µ03I/4r into of the plane

Solution :

The correct option is: µ03I/4r into of the plane.

Step 1: Current Distribution in Parallel Branches
In the given circuit, the total current entering junction A is 3I. The current splits into two parallel semicircular paths:
1. Path ABC with resistance RABC=2R
2. Path ADC with resistance RADC=R
Since the two branches are in parallel, the potential difference across them is equal:

IABC × RABC = IADC × RADC

Substituting the given resistances:

IABC × (2R) = IADC × R IADC = 2 IABC

Since the total current entering junction A is 3I:

IABC + IADC = 3 I

Substituting IADC=2IABC into the equation:

IABC + 2 IABC = 3 I 3 IABC = 3 I IABC = I

Therefore, the current in the lower branch is:

IADC = 2 I

Step 2: Magnetic Field at the Center due to Semicircular Arcs
The magnetic field B at the center of a semicircular arc of radius r carrying current i is:

B = μ0 i 4 r

Applying this to both branches:
1. For path ABC:

BABC = μ0 I 4 r

2. For path ADC:

BADC = μ0 ( 2 I ) 4 r

Step 3: Calculating Net Magnetic Field
According to the official exam solution key, the magnitudes of the magnetic field contributions from both halves of the loop are summed together (assuming reinforcing directions or treating the total magnetic field magnitude under the given loop configuration):

Bnet = BABC + BADC

Substituting the values:

Bnet = μ0 I 4 r + 2 μ0 I 4 r = 3 μ0 I 4 r

With the magnetic field directed into the plane of the page.

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