Two parallel wires Q and R carry currents 5 A and 10 A respectively in opposite directions. A plane lamina ABCD intersects the wires at right angles at points Q and R. Find the magnitude of the total magnetic induction at point P located in the lamina as shown in the figure.

Solution

Given,
Current carried by wire Q, I₁ = 5 A
Current carried by wire R, I₂ = 10 A

The magnetic induction at P due to the 10 A current, is

B_{1} = \frac{μ_{0} I_{1}}{2 {πr}_{1}} = \frac{4 π \times 10^{- 7} \times 10}{2 π \times 200 \times 10^{- 3}} T = 10 μT

The magnetic induction at P due to the 5 A current,

B_{2} = \frac{μ_{0} I_{2}}{2 {πr}_{2}}

B_{2} = \frac{4 A \times 10^{- 7} \times 5}{2 A \times 80 \times 10^{- 3}} B_{2} = \frac{5 \times 100 \times 10^{- 7}}{100} = 125 \times 10^{- 1} \times 10^{- 6} = 12.5 \times μT

Hence, Resultant magnetic induction is,

B = \sqrt{B_{1}^{2} + B_{2}^{2} + 2 B_{1} B_{2} \cos (180 ° - θ)}

where,

{QR}^{2} = {PQ}^{2} + {PR}^{2} - 2 PQ . PR \cos θ

\cos θ = \frac{- 150^{2} + 80^{2} + 200^{2}}{2 \times 80 \times 200} = + 0.7469

Therefore,

B = 10^{- 6} \sqrt{10^{2} + 12 . 5^{2} - 2 \times 10 \times 12.5 \times 0.7469} = 8.34 \times 10^{- 6} T = 8.34 μT

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