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 ________.
Correct Answer :
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 . The current splits into two parallel semicircular paths:
1. Path ABC with resistance
2. Path ADC with resistance
Since the two branches are in parallel, the potential difference across them is equal:
Substituting the given resistances:
Since the total current entering junction A is :
Substituting into the equation:
Therefore, the current in the lower branch is:
Step 2: Magnetic Field at the Center due to Semicircular Arcs
The magnetic field at the center of a semicircular arc of radius carrying current is:
Applying this to both branches:
1. For path ABC:
2. For path ADC:
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):
Substituting the values:
With the magnetic field directed into the plane of the page.
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