A player with a 3 text{ m} long iron rod runs | vedclass

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(B) The induced electromotive force (emf) in a conductor of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by $e = Bvl \sin 	he

Vedclass · Accepted Answer

$1 	imes 10^{-3} 	ext{ V}$ in the vertical position and zero in the horizontal position

Vedclass · Answer

(B) The induced electromotive force (emf) in a conductor of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by $e = Bvl \sin 	heta$,where $	heta$ is the angle between the velocity vector and the magnetic field vector. The Earth's horizontal magnetic field $B_H$ acts from South to North.$1$. When the rod is held vertically,the rod is perpendicular to both the velocity (East) and the magnetic field (North). Thus,the rod cuts the magnetic field lines. The induced emf is:$v = 30 	ext{ km/hr} = 30 	imes \frac{5}{18} 	ext{ m/s} = \frac{25}{3} 	ext{ m/s}$$e = B_H \cdot v \cdot l = (4 	imes 10^{-5} 	ext{ Wb/m}^2) 	imes (\frac{25}{3} 	ext{ m/s}) 	imes (3 	ext{ m}) = 1 	imes 10^{-3} 	ext{ V}$.$2$. When the rod is held horizontally and parallel to the direction of motion (East),the rod is parallel to the velocity vector. Since the rod is not cutting the magnetic field lines,the induced emf is $e = 0$.

Similar Questions

$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with a uniform speed $v = 20\, m/s$ in a uniform magnetic field $B = 4.0\, T$ directed into the paper. $A$ capacitor of capacity $C = 10\, \mu F$ is connected as shown in the figure. Then:

$A$ metallic rod of length $1 \,m$ held along the east-west direction is allowed to fall down freely. Given the horizontal component of the Earth's magnetic field $B_H = 3 \times 10^{-5} \,T$. The emf induced in the rod at an instant $t = 2 \,s$ after it is released is (Take $g = 10 \,m/s^2$):

$A$ $10 \, m$ 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:

$A$ straight conductor of length $4 \,m$ moves at a speed of $10 \,m/s$. When the conductor makes an angle of $30^{\circ}$ with the direction of a magnetic field of induction $0.1 \,Wb/m^2$, the induced emf is: (in $\,V$)

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