A conducting rod of length ell is moving with | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To determine the potential difference across the moving conducting rod,we use the Lorentz force law on the free electrons within the rod.According

Vedclass · Accepted Answer

$Q$

Vedclass · Answer

(B) To determine the potential difference across the moving conducting rod,we use the Lorentz force law on the free electrons within the rod.According to the right-hand rule for the motional electromotive force $(EMF)$,the force $F$ on a positive charge $q$ moving with velocity $v$ in a magnetic field $B$ is given by $F = q(v 	imes B)$.For electrons (charge $q = -e$),the force is $F = -e(v 	imes B)$.Using the right-hand rule for $(v 	imes B)$,the direction of $(v 	imes B)$ is towards the end $P$. Therefore,the force on the electrons is directed towards the end $Q$.As electrons accumulate at end $Q$,it becomes negatively charged (lower potential),and end $P$ becomes positively charged (higher potential).Thus,end $Q$ is at a lower potential.

$A$ conducting rod of length $\ell$ is moving with velocity $v$ in a uniform magnetic field $B$ directed into the plane of the paper,as shown in the diagram. Which end of the rod is at a lower potential?

Similar Questions

If ${\left( \frac{2}{3} \right)^{x + 2}} = {\left( \frac{3}{2} \right)^{2 - 2x}}$,then $x =$

What is $Y$ in the following reaction?
$C_2H_5I + NaOC_2H_5 \longrightarrow X + NaI$
$X + 2HI \xrightarrow{\Delta} 2Y + H_2O$

Which of the following compounds is soluble in benzene but almost insoluble in water?

$A$ pair of perpendicular lines passes through the origin and also through the points of intersection of the curve $x^2+y^2=4$ with $x+y=a$,where $a>0$. Then $a$ is equal to

In a fission reaction $_{92}^{236}U \to ^{117}X + ^{117}Y + n + n$,the binding energy per nucleon of $X$ and $Y$ is $8.5\, MeV$,whereas that of $^{236}U$ is $7.6\, MeV$. The total energy liberated will be about:

Vedclass Test Series

Exam Paper Generator

Online Exam Module