Questions
The magnetic field at center ‘O’ of the arc in figure is
a
μoI4π×r2+π
b
μoI2πrπ4+2−1
c
μ04π×Ir2−π
d
μ04π×Ir2+π4
detailed solution
Correct option is B
Magnetic field due to straight wires and quarter circle is given byBcircular=14μ0I2r=μ0I8r Bstraight=2μ0I4π(rcos450)sin900−sin450 =2μ0I2πr1−12 Hence, Bnet = μ0I8r+2μ0I2πr1−12 = μoI2πrπ4+2−1Talk to our academic expert!
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