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Physics · 2025 · 3 marks
Moving Charges and MagnetismMagnetic Field due to a Current Element, Biot-Savart Law3 marksApplyshort answer
CBSE 2025 · Region 7 · Set 1 · Q23
(a) (i) Write Biot-Savart's law in vector form. (ii) Two identical circular coils A and B, each of radius R, carrying currents I and √ 3 I…
(i)
Write Biot-Savart's law in vector form.
(ii)
Two identical circular coils A and B , each of radius R , carrying currents $\displaystyle I$ and $\displaystyle \sqrt{3} I$ respectively, are placed concentrically in XY and YZ planes respectively. Find the magnitude and direction of the net magnetic field at their common centre.
(i)
A rectangular loop of sides $\displaystyle l$ and b carries a current I clockwise. Write the magnetic moment $\displaystyle \overrightarrow{\mathrm{m}}$ of the loop and show its direction in a diagram.
(ii)
The loop is placed in a uniform magnetic field $\displaystyle \overrightarrow{\mathrm{B}}$ and is free to rotate about an axis which is perpendicular to $\displaystyle \overrightarrow{\mathrm{B}}$. Prove that the loop experiences no net force, but a torque $\displaystyle \vec{\tau}=\overrightarrow{\mathrm{m}} \times \overrightarrow{\mathrm{B}}$.
Marking-scheme solution
(i)
$\displaystyle \overrightarrow{dB} = \dfrac{\mu_0}{4\pi}\dfrac{I(\overrightarrow{dl}\times\overrightarrow{r})}{r^3}$
(ii)
$\displaystyle B_1 = \dfrac{\mu_0 I}{2R}$
$\displaystyle B_2 = \dfrac{\mu_0\sqrt{3}I}{2R}$
$\displaystyle B = \sqrt{B_1^2 + B_2^2}$
$\displaystyle \therefore B = \dfrac{\mu_0 I}{2R}\sqrt{1+3}$
$\displaystyle B = \dfrac{\mu_0 I}{R}$
$\displaystyle \tan\theta = \dfrac{B_1}{B_2} = \dfrac{1}{\sqrt{3}}$
Direction of net magnetic field is $\displaystyle 30$° with direction of $\displaystyle B_2$ / $\displaystyle 60$° with the direction of $\displaystyle B_1$.
(i)
$\displaystyle \vec{m} = I\vec{A}$
(ii)
$\displaystyle F_1 = F_2 = IbB$ $\displaystyle F_1 =$ Force on AB into the plane
$\displaystyle F_2 =$ Force on CD out of the plane
Since forces are equal & opposite so net force = $\displaystyle 0$
Both forces form a couple, magnitude of torque acting on the coil is
$\displaystyle \therefore \tau = F_1\dfrac{l}{2}\sin\theta + F_2\dfrac{l}{2}\sin\theta$
$\displaystyle = I\,b\,B\,l\sin\theta$
$\displaystyle = I\,A\,B\,\sin\theta$
$\displaystyle = m\,B\sin\theta$
$\displaystyle \vec{\tau} = \vec{m}\times\vec{B}$
Alternatively:
If the plane of the current carrying coil makes an angle $\displaystyle \propto$ with the magnetic field
$\displaystyle \vec{F}_{DA} = -\vec{F}_{Bc}$ (cancel each other)
Force on the arm DC is into the plane of the paper
$\displaystyle \left|F_{DC}\right| = IbB$
Force on the arm AB is out of the plane of the paper.
$\displaystyle \left|F_{AB}\right| = IbB$
Since forces are equal & opposite so net force = $\displaystyle 0$
Both of them form a couple and magnitude of torque acting on the coil is
$\displaystyle \tau$ = either force × perpendicular distance between the two forces.
$\displaystyle \tau = IbB \times l\sin\theta$
$\displaystyle = IAB\sin\theta$
$\displaystyle \vec{\tau} = I\vec{A}\times\vec{B}$
$\displaystyle \vec{\tau} = \vec{m}\times\vec{B}$
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