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CBSE Class 12 Physics 2017 Outside Delhi Set 2 Paper

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Question : 9 of 9
Marks: +1, -0
Two identical loops PP and QQ each of radius 5 cm5\,\text{cm} are lying in perpendicular planes such that they have a common centre as shown in the figure. Find the magnitude and direction of the net magnetic field at the common centre of the two coils, if they carry currents equal to 3 A3\,\text{A} and 4 A4\,\text{A} respectively.
Solution:  
Formula
Field due to each coil
Magnitude of resultant field
Direction of resultant field
Field at the centre of a circular coil =μ0I2R=\frac{\mu_0 I}{2 R}
Field due to coil P=μ0×32×5×10−2P=\frac{\mu_0 \times 3}{2 \times 5 \times 10^{-2}} tesla
=12π×10−6 tesla=12\pi \times 10^{-6}\,\text{tesla}
Field due to coil Q=μ0×32×5×10−2Q=\frac{\mu_0 \times 3}{2 \times 5 \times 10^{-2}} tesla
=16π×10−6 tesla=16\pi \times 10^{-6}\,\text{tesla}
∴\therefore Resultant Field =(π122+162) μT= (\pi \sqrt{12^2+16^2})\,\mu T
=(20π) μT= (20\pi)\,\mu T
Let the field make an angle θ\theta with the vertical
tan⁡θ=12π×10−616π×10−6=34\tan\theta=\frac{12\pi \times 10^{-6}}{16\pi \times 10^{-6}}=\frac{3}{4}
θ=tan⁡−134\theta=\tan^{-1}\frac{3}{4}
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