The magnetic field in a region is given by ov | vedclass

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

Question

(C) The motional emf induced in a conductor of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by $E = B l v$.The square loop has

Vedclass · Accepted Answer

$\frac{ B _{0} v _{0} d ^{2}}{ a }$

Vedclass · Answer

(C) The motional emf induced in a conductor of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by $E = B l v$.The square loop has two vertical sides of length $d$ at positions $x$ and $x+d$.The emf induced in the side at $x+d$ is $E _{1} = B(x+d) \cdot d \cdot v _{0} = B _{0} \left( \frac{x+d}{a} \right) d v _{0}$.The emf induced in the side at $x$ is $E _{2} = B(x) \cdot d \cdot v _{0} = B _{0} \left( \frac{x}{a} \right) d v _{0}$.The net induced emf in the loop is the difference between these two emfs: $E _{net} = E _{1} - E _{2}$.$E _{net} = \frac{B _{0} d v _{0}}{a} (x+d - x) = \frac{B _{0} v _{0} d ^{2}}{a}$.

Similar Questions

$A$ conducting ring of radius $R$ is placed in a uniform inward magnetic field $\vec B$ as shown. If the ring is moving with velocity $\vec v$ in its plane,the induced $emf$ across the arc $PQ$ will be:

$A$ conducting wire is moving towards the right in a magnetic field $B$. The direction of the induced current $i$ in the wire is shown in the figure. The direction of the magnetic field will be

$A$ long solenoid with $2000$ turns per meter has a small loop of radius $3 \,cm$ placed inside the solenoid normal to its axis. If the current through the solenoid increases steadily from $1.5 \,A$ to $5.5 \,A$ in $\frac{\pi^2}{100} \,s$, the induced emf in the loop is (in $\,mV$)

$A$ $10 \, m$ wire kept in the east-west direction is falling with a velocity of $5 \, m/s$ perpendicular to the magnetic field of $0.3 \times 10^{-4} \, Wb/m^2$. The induced e.m.f. across the terminals of the wire will be:

$A$ jet plane of wing span $20 \,m$ is travelling towards west at a speed of $400 \,ms^{-1}$. If the earth's total magnetic field is $4 \times 10^{-4} \,T$ and the dip angle is $30^{\circ}$ at that place, the voltage difference developed across the ends of the wing is (in $\,V$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module