14 questions · self-marked practice — reveal the answer and mark yourself.
Force due to the wire AB and force due to wire CD are equal and opposite to each other.
Thus they cancel each other.
Net force is the force due to the semicircular loop = 2iRB
$\text{U}=-\vec{\text{m}}.\vec{\text{B}}$
In the case of stable equilibrium potential energy is minimum. So, far $\theta=0^\circ$ Potential Energy is -ve and minimum.$\Rightarrow\text{d}\phi=\text{dt}$
Charge in flux has unit weber and potential difference as volt.
Force on a semicircular wire
= 2iRB
= 2 × 5 × 0.05 × 0.5
= 0.25N
Let ‘F’ be the force applied due to magnetic field on the wire and ‘x’ be the dist covered.
So, $\text{F}\times\text{l}=\mu\text{mg}\times\text{x}$
$\Rightarrow\text{ibBl}=\mu\text{mgx}$
$\Rightarrow\text{x}\frac{\text{ibBl}}{\mu\text{mg}}$
$\text{i}=10\text{A},\ \text{B}=0.1\text{T},\ \theta=53^\circ$
$|\text{F}|=\text{iL B}\sin\theta=10\times10^{-1}\times0.1\times0.79$
$=0.0798\approx0.08$
direction of F is along a direction $\perp\text{r}$ to both l and B.
Radius = l, K.E = K
$\text{L}=\frac{\text{Mv}}{\text{qB}}$
$\Rightarrow\text{l}=\frac{\sqrt{2\text{mk}}}{\text{ql}}$
$\Rightarrow\text{B}=\sqrt{\frac{2\text{mk}}{\text{ql}}}$
$\text{r}=0.02\text{m}$
$\theta=30^\circ$
$\text{i}=2\text{A}$
$\text{B}=4\times10^{-1}\text{T}$
$=500\times1\times3.14\times4\times10^{-4}\times4\times10^{-1}\times\Big(\frac{1}{2}\Big)$
$12.56\times10^{-2}=0.1256\approx0.13\text{N-M}$
$\text{v}=3\times10^4\text{km/s}$
$\text{B}=1\text{T},\ \text{F}=\text{qBv}$
$=2\times1.6\times10^{19}\times3\times10^7\times1$
$=9.610^{12}\text{N}.$
$\Rightarrow\tau=\text{ni}\text{ AB}\sin90^\circ$
$\Rightarrow0.2=100\times2\times5\times4\times10^{-4}\times\text{B}$
$\Rightarrow\text{B}=\frac{0.2}{100\times2\times5\times4\times10^{-4}}=0.5$ Tesla