Question:

Two wires AA and BB are of lengths 40cm 40\,cm and 30cm 30\,cm . AA is bent into a circle of radius rr and BB into an arc of radius rr. A current i1i_{1} is passed through AA and I2I_2 through BB. To have the same magnetic inductions at the centre, the ratio of I1:I2I_{1}:I_{2} is

Updated On: Jun 2, 2024
  • 3:4 3:4
  • 3:5 3:5
  • 2:3 2:3
  • 4:3 4:3
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The Correct Option is A

Solution and Explanation

For wire A Length =40cm=40\, cm
It is bent into a circle, so
2πr=40\therefore 2 \pi r=40
r=402πr=\frac{40}{2 \pi}
Magnetic induction at centre =μ0i12r=\frac{\mu_{0} i_{1}}{2 r}
For wire B Length =30cm=30 cm It is bent into an arc, so
θr=30\therefore \theta r=30
Magnetic induction at centre due to circular arc
=μ0i2θ4πr=\frac{\mu_{0} i_{2} \theta}{4 \pi r}
Since, μ0i12r=μ0i1θ4πr\frac{\mu_{0} i_{1}}{2 r}=\frac{\mu_{0} i_{1} \theta}{4 \pi r}
i1i2=θ2π=30×2π2π×40\Rightarrow \frac{i_{1}}{i_{2}}=\frac{\theta}{2 \pi}=\frac{30 \times 2 \pi}{2 \pi \times 40}
i1i2=34\Rightarrow \frac{i_{1}}{i_{2}}=\frac{3}{4}
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Magnetic Field:

Region in space around a magnet where the Magnet has its Magnetic effect is called the Magnetic field of the Magnet. Let us suppose that there is a point charge q (moving with a velocity v and, located at r at a given time t) in presence of both the electric field E (r) and the magnetic field B (r). The force on an electric charge q due to both of them can be written as,

F = q [ E (r) + v × B (r)] ≡ EElectric +Fmagnetic 

This force was based on the extensive experiments of Ampere and others. It is called the Lorentz force.