LEVELJEE Main

Animated Solution for Physics - Electromagnetic Induction: A metal rod moves at a constant velocity in a direction perpendicular to its length. A constant uniform magnetic field exists in space in a direction perpendicular to the rod as well as its velocity. Select the correct statement (s) from the following.

Q: Does the electric field keep increasing indefinitely?

No, it increases only until the electric force $qE$ exactly balances the magnetic force $qvB$. After that, a steady state is reached.

Q: How does the potential vary along the rod?

In steady state, the electric field $E = vB$ is uniform. Since $E = -dV/dx$, the potential varies linearly along the length of the rod.

Select Answer:

Visualized Solution

Visualizing the Setup

Let the rod $A B$ move with velocity $v$ in a uniform magnetic field $B$ .

Lorentz Force on Electrons

Free electrons in the rod experience a magnetic Lorentz force:
$F_{m} = - e (v \times B)$

Charge Accumulation

By Fleming's Left Hand Rule, electrons are pushed towards end $B$ .
End $B$ becomes negatively charged, and end $A$ becomes positively charged.

Induced Electric Field

The separation of charges creates a potential difference $V_{A} - V_{B} > 0$ .
This establishes an internal electric field $E$ from $A$ to $B$ .

Conclusion

Since an electric field exists, the potential varies linearly along the rod.
Therefore, there is an electric field in the rod.

Steady State Equilibrium

In steady state, electric force balances magnetic force:
$e E = e v B ⟹ E = v B$
Motional EMF: $ε = v B l$

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The Sigma Insight: Motional EMF

Solution Diagram

Analyzing the Setup Imagine a metal rod moving smoothly with a constant velocity through a uniform magnetic field

The magnetic field is perpendicular to both the length of the rod and its direction of motion.

Inside this metal rod, there are countless free electrons. As the rod moves, these electrons are also carried along through the magnetic field. Because they are moving charges in a magnetic field, they experience a magnetic Lorentz force given by:

F_{m} = - e (v \times B)

The Charge Separation Using the right-hand rule for the cross product of velocity and magnetic field, we find the force on positive charges is towards one end, say end $A$

Consequently, the free electrons are pushed towards the opposite end,

B

This causes a buildup of negative charge at

B

and positive charge at

A

. This separation of charges creates a potential difference between the two ends of the rod. End

A

becomes positively charged, and end

B

becomes negatively charged.

The Internal Electric Field

What does a potential difference imply? It means an internal electric field

E

is established inside the rod, directed from the positive end to the negative end.

Because of this internal electric field, the entire rod cannot possibly be at the same electric potential. The potential varies linearly from one end to the other. Thus, the correct statement is that there is an electric field present inside the rod.

Steady State Equilibrium Think about what happens eventually in a steady state

The electric field will grow until the electric force exactly balances the magnetic force on the electrons:

e E = e v B ⟹ E = v B

At this point, the charge accumulation stops, leading to a steady motional EMF across the length

l

of the rod:

ε = v B l

Common Traps & FAQs

Why isn't the entire rod at the same potential?

Because the magnetic force pushes charges to opposite ends, creating a potential difference. A conductor is only an equipotential volume in electrostatics, not when it's moving in a magnetic field.

Does the electric field keep increasing indefinitely?

No, it increases only until the electric force

q E

exactly balances the magnetic force

q v B

. After that, a steady state is reached.

What is the direction of the electric field?

The electric field is directed from the positively charged end to the negatively charged end, which opposes the magnetic force on the electrons.

Is there a current flowing in the rod?

There is a transient current while the charges are accumulating. Once steady state is reached, the net force on charges is zero, and no continuous current flows unless the rod is part of a closed circuit.

How does the potential vary along the rod?

In steady state, the electric field

E = v B

is uniform. Since

E = - d V / d x

, the potential varies linearly along the length of the rod.

Similar Questions

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A metallic rod of length $l$ is tied to a string of length $2 l$ and made to rotate with angular speed $ω$ on a horizontal table with one end of the string fixed. If there is a vertical magnetic field $B$ in the region, the emf induced across the ends of the rod is

(A)

\frac{2 B ω l ^{2}}{2}

(B)

\frac{3 B ω l ^{2}}{2}

(C)

\frac{4 B ω l ^{2}}{2}

(D)

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A conducting bar of length $L$ is free to slide on two parallel conducting rails as shown in the figure. Two resistors $R_{1}$ and $R_{2}$ are connected across the ends of the rails. There is a uniform magnetic field $B$ pointing into the page. An external agent pulls the bar to the left at a constant speed $v$ . The correct statement about the directions of induced currents $I_{1}$ and $I_{2}$ flowing through $R_{1}$ and $R_{2}$ respectively is

(A)

both

I_{1}

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both

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A long straight wire carries a current, $I = 2$ ampere. A semi-circular conducting rod is placed beside it on two conducting parallel rails of negligible resistance. Both the rails are parallel to the wire. The wire, the rod and the rails lie in the same horizontal plane, as shown in the figure. Two ends of the semi-circular rod are at distances $1 cm$ and $4 cm$ from the wire. At time $t = 0$ , the rod starts moving on the rails with a speed $v = 3.0 m/s$ (see the figure). A resistor $R = 1.4 Ω$ and a capacitor $C_{0} = 5.0 μ F$ are connected in series between the rails. At time $t = 0$ , $C_{0}$ is uncharged. Which of the following statement(s) is(are) correct ? [ $μ_{0} = 4 π \times 1 0^{- 7}$ SI units. Take $ln 2 = 0.7$ ]

* Multiple Correct Options

(A)

Maximum current through

R

1.2 \times 1 0^{- 6} ampere

(B)

Maximum current through

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(C)

Maximum charge on capacitor

C_{0}

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(D)

Maximum charge on capacitor

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A metallic square loop $A B C D$ is moving in its own plane with velocity $v$ in a uniform magnetic field perpendicular to its plane as shown in the figure. Electric field is induced

(A)

A D

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(B)

B C

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(C)

neither in

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(D)

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A conducting wire of parabolic shape, initially $y = x^{2}$ , is moving with velocity $V = V_{0} \hat{i}$ in a non-uniform magnetic field $B = B_{0} (1 + (\frac{y}{L})^{β}) \hat{k}$ , as shown in figure. If $V_{0}, B_{0}, L$ and $β$ are positive constants and $Δ ϕ$ is the potential difference developed between the ends of the wire, then the correct statement(s) is/are:

* Multiple Correct Options

(A)

∣Δ ϕ ∣

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y = x

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(B)

∣Δ ϕ ∣

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(C)

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∣Δ ϕ ∣ = \frac{4}{3} B_{0} V_{0} L

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Two long parallel horizontal rails, a distance $d$ apart and each having a resistance $λ$ per unit length, are joined at one end by a resistance $R$ . A perfectly conducting rod $M N$ of mass $m$ is free to slide along the rails without friction (see figure). There is a uniform magnetic field of induction $B$ normal to the plane of the paper and directed into the paper. A variable force $F$ is applied to the rod $M N$ such that, as the rod moves, a constant current $i$ flows through $R$ . (i) Find the velocity of the rod and the applied force $F$ as functions of the distance $x$ of the rod from $R$ . (ii) What fraction of the work done per second is converted into heat.

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A conducting square loop of side $L$ and resistance $R$ moves in its plane with a uniform velocity $v$ perpendicular to one of its sides. A magnetic induction $B$ constant in time and space, pointing perpendicular and into the plane at the loop exists everywhere with half the loop outside the field, as shown in figure. The induced emf is

(A)

zero

(B)

R v B

(C)

\frac{v B L}{R}

(D)

v B L

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A metal conductor of length $1 m$ rotates vertically about one of its ends at angular velocity $5 rad/s$ . If the horizontal component of earth's magnetic field is $0.2 \times 1 0^{- 4} T$ , then the emf developed between the two ends of the conductor is

(A)

5 μ V

(B)

50 μ V

(C)

5 mV

(D)

50 mV

JEE Advanced 1996

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A thin semicircular conducting ring of radius $R$ is falling with its plane vertical in a horizontal magnetic induction $B$ . At the position $M N Q$ the speed of the ring is $v$ and the potential difference developed across the ring is

(A)

zero

(B)

B v π R^{2} /2

and

M

is at higher potential

(C)

π B R v

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Q

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(D)

2 R B v

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A constant magnetic field of $1 T$ is applied in the $x > 0$ region. A metallic circular ring of radius $1 m$ is moving with a constant velocity of $1 m/s$ along the X-axis. At $t = 0 s$ , the centre of O of the ring is at $x = - 1 m$ . What will be the value of the induced emf in the ring at $t = 1 s$ ? (Assume the velocity of the ring does not change.)

(A)

1 V

(B)

2 π V

(C)

2 V

(D)

0

Select Answer:

Visualized Solution

Visualizing the Setup

Lorentz Force on Electrons

Charge Accumulation

Induced Electric Field

Conclusion

Steady State Equilibrium

The Sigma Insight: Motional EMF

Analyzing the Setup Imagine a metal rod moving smoothly with a constant velocity through a uniform magnetic field

The Charge Separation Using the right-hand rule for the cross product of velocity and magnetic field, we find the force on positive charges is towards one end, say end .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

The Internal Electric Field

Steady State Equilibrium Think about what happens eventually in a steady state

Similar Questions

The Charge Separation Using the right-hand rule for the cross product of velocity and magnetic field, we find the force on positive charges is towards one end, say end $A$