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Physics · 2025 · 5 marks
CBSE 2025 · Region 1 · Set 2 · Q33
(a) (i) A proton moving with velocity V in a non-uniform magnetic field traces a path as shown in the figure. The path followed by the proton is…
(i)
A proton moving with velocity $\displaystyle \overrightarrow{\mathrm{V}}$ in a non-uniform magnetic field traces a path as shown in the figure.
The path followed by the proton is always in the plane of the paper. What is the direction of the magnetic field in the region near points $\displaystyle \mathrm{P}, \mathrm{Q}$ and R ? What can you say about relative magnitude of magnetic fields at these points?
(ii)
A current carrying circular loop of area A produces a magnetic field B at its centre. Show that the magnetic moment of the loop is $\displaystyle \frac{2 \mathrm{BA}}{\mu_{0}} \sqrt{\frac{\mathrm{~A}}{\pi}}$.
(i)
Derive an expression for the torque acting on a rectangular current loop suspended in a uniform magnetic field.
(ii)
A charged particle is moving in a circular path with velocity $\displaystyle \vec{\mathrm{V}}$ in a uniform magnetic field $\displaystyle \overrightarrow{\mathrm{B}}$. It is made to pass through a sheet of lead and as a consequence, it looses one half of its kinetic energy without change in its direction. How will ($\displaystyle 1$) the radius of its path
(2)
its time period of revolution change ?
Marking-scheme solution
(a)
Near point P: magnetic field is acting into the plane of the paper as force is acting upwards.
Near point Q: magnetic field is into the plane of paper as force is acting upwards.
Near point R: magnetic field is acting out of the plane of the paper as force is acting downwards.
Relative magnitude of the magnetic field:
As $\displaystyle B \propto \frac{1}{r}$
Therefore, near point P, magnitude of B is small.
Near point Q, B is relatively smaller than point P.
Near point R, B is relatively larger than point P.
$\displaystyle \left(B_{Q}<B_{P}<B_{R}\right)$
(ii)
Let r be the radius of the circular coil and I the current in the coil, then:
$\displaystyle B=\frac{\mu_{0} I}{2 r}$ or $\displaystyle I=\frac{2 B r}{\mu_{0}}$
$\displaystyle A=\pi r^{2} \quad r=\sqrt{\frac{A}{\pi}}$
$\displaystyle M=I A$
$\displaystyle =\frac{2 B r}{\mu_{0}} A$
$\displaystyle =\frac{2 B A}{\mu_{0}} \sqrt{\frac{A}{\pi}}$
(b)
$\displaystyle F_{1}$ and $\displaystyle F_{2}$ are the forces acting on two arms of the rectangular coil having sides a and b.
$\displaystyle F_{1}=F_{2}=I b B$ (b = length of the arm)
Forces constitute a couple. The magnitude of torque on the loop is:
$\displaystyle \tau=F_{1} \frac{a}{2} \sin \theta+F_{2} \frac{a}{2} \sin \theta$
$\displaystyle =I a b B \sin \theta$
$\displaystyle =I A B \sin \theta$
$\displaystyle \vec{\tau}=I \vec{A} \times \vec{B}$
If the plane of the current carrying coil makes an angle $\displaystyle \alpha$ with the magnetic field:
$\displaystyle \vec{F}_{D A}=-\vec{F}_{B C}$ (cancel each other)
Force on the arm DC is into the plane of the paper, $\displaystyle F_{D C}=I b B$.
Force on the arm AB is out of the plane of the paper, $\displaystyle F_{A B}=I b B$.
Both of them form a couple and torque acting on the coil is:
$\displaystyle \tau=$ either force $\displaystyle \times$ perpendicular distance between the two forces
$\displaystyle \tau=I b B \times a \cos \alpha$
$\displaystyle =I a b B \cos \alpha$
$\displaystyle \tau=I A B \cos \alpha$
Let $\displaystyle \hat{n}=$ outward drawn normal to the plane of the coil.
$\displaystyle \theta+\alpha=90^{\circ}$
$\displaystyle \alpha=90^{\circ}-\theta$
$\displaystyle \tau=I A B \cos (90-\theta)$
$\displaystyle =I A B \sin \theta$
$\displaystyle \vec{\tau}=I \vec{A} \times \vec{B}$
(ii)(1)
$\displaystyle r=\frac{m v}{q B}=\frac{\sqrt{2 m K}}{q B}$
$\displaystyle r \propto \sqrt{K}$
$\displaystyle \frac{r^{\prime}}{r}=\sqrt{\frac{K / 2}{K}}=\frac{1}{\sqrt{2}}$
$\displaystyle r^{\prime}=\frac{r}{\sqrt{2}}$
(2)
$\displaystyle T=\frac{2 \pi m}{q B}$
Time period does not depend on kinetic energy.
$\displaystyle \therefore$ Time period will not change.
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