moving-charges-and-magnetism Question 12
Question: Q. 4. Two identical loops $P$ and $Q$ each of radius $5 \mathrm{~cm}$ 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 $3 \mathrm{~A}$ and $4 \mathrm{~A}$ respectively.
$$ \begin{aligned} & \text {
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Solution:
Ans. Formula 11/2 } \ & \text { Field due to each coil } \ & \text { Magnitude of resultant field } \ & \text { Direction of resultant field } \ & 1 / 2+1 / 2 \ & \text { Field at the centre of a circular coil }=\frac{\mu_{0} I}{2 R} \ & =12 \pi \times 10^{-6} \text { tesla } \quad 1 / 2 \ & =16 \pi \times 10^{-6} \text { tesla } \quad 1 / 2 \ & =(20 \pi) \mu \mathrm{T} \end{aligned} $$
Let the field make an angle $\theta$ with the vertical
$$ \begin{aligned} \tan \theta & =\frac{12 \pi \times 10^{-6}}{16 \pi \times 10^{-6}}=\frac{3}{4} \ \theta & =\tan ^{-1} \frac{3}{4} \end{aligned} $$
$1 / 2$
(Alternatively : $\theta^{\prime}=\tan ^{-1} \frac{4}{3}, \theta^{\prime}=$ angle with the horizontal)
[Note 1 : Award 2 marks if the student directly calculates $B$ without calculating $B_{P}$ and $B_{Q}$ separately.]
[Note 2 : Some students may calculate the field $B_{Q}$ and state that it also represents the resultant magnetic field (as coil $P$ has been shown ‘broken’ and therefore, cannot produce a magnetic field); They may be given $2 \frac{1}{2}$ marks for their (correct) calculation of $\left.B_{Q}\right]$
[CBSE Marking Scheme 2017]
Detailed Answer :
Try yourself, Similar to Q. 3, Short Answer Type I
