- (b) Fleming's right hand rule.
Explanation:
Direction of current induced in a wire moving in a magnetic field is found by using Fleming's right hand rule.
- (a) $\frac{\text{Blv}}{R}$
Explanation:
Induced e.m.f $\varepsilon=\text{Blv}$
Current in the rod, $\text{l}=\frac{\varepsilon}{\text{R}}=\frac{\text{Blv}}{\text{R}}$
- (c) 6.25 mW
Explanation:
Here, $\text{l}=0.1\text{m,}\text{v}=1\text{ms}^{-1}$
$\text{I}=50\text{A, B}=1.25\text{mT}=1.25\times10^{-3}\text{T}$
The induced emf is, $\varepsilon=\text{Blv}$
The mechanical power is,
$\text{P}=\varepsilon\text{I}=\text{BlvI}=1.25\times10^{-3}\times0.1\times1\times50$
$=6.25\times10^{-3}\text{W}=6.25\text{mW}$
- (d) 1 volt
Explanation:
Emf induced, $\varepsilon=\text{Blv}$
Here, $\vec{\text{B}},\vec{\text{l}}$ and $\vec{\text{v}}$ are mutually perpendicular For given B and $\text{l},\varepsilon\propto\text{v}.$
$\therefore\frac{\varepsilon_1}{\varepsilon_2}=\frac{\text{v}_1}{\text{v}_2}$
Here, $\varepsilon_1=1.5\text{V},\text{v}_1=15\text{km/ hr}=15\times\frac{5}{18}\text{ms}^{-1}$
$\text{v}_210\text{km/ hr}=10\times\frac{5}{18}\text{m s}^{-1},\varepsilon_2=?$
$\frac{1.5}{\varepsilon_2}=\frac{15\frac{5}{18}}{10\times\frac{5}{18}}=\frac{3}{2};\varepsilon_2=1\text{V}$
- (a) [ML2T-3A-1]
Explanation:
$\varepsilon=\frac{[\text{W}]}{[\text{q}]}=\frac{\text{ML}^2\text{T}^{-2}}{\text{AT}}=\text{ML}^2\text{T}^{-3}\text{A}^{-1}$