Solve this.
Q.11. For the arrangement shown in figure, the magnetic induction at the centre C


(1) 3 μ 0 I 16 R
(2) μ 0 I 16 R
(3) 5 μ 0 I 16 R
(4) 16 μ 0 I 5 R

Dear Student,
                      In the figure there are two arcs larger one subtends  3π2 on the centre and smaller one π2So net magnetic field at the centre=μ0I2×2R×34-μ0I2R×14=μ0I16R
Regards

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