Q35P The conducting rod shown in Figu... [FREE SOLUTION]

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Chapter 30: Q35P (page 898)

The conducting rod shown in Figure has length L and is being pulled along horizontal, frictionless conducting rails at a constant velocity $\overset{鈫�}{v}$ . The rails are connected at one end with a metal strip. A uniform magnetic field $\overset{鈫�}{B}$ , directed out of the page, fills the region in which the rod moves. Assume that $L = 10 c m, v = 5.0 \frac{m}{s}, a n d B = 1.2 T$ . (a) What is the magnitude of the emf induced in the rod? (b) What is the direction (up or down the page) of the emf induced in the rod? (c) What is the size of the current in the conducting loop? (d) What is the direction of the current in the conducting loop? Assume that the resistance of the rod is 0.40 and that the resistance of the rails and metal strip is negligibly small. (e) At what rate is thermal energy being generated in the rod? (f) What external force on the rod is needed to maintain $\overset{鈫�}{v}$ ? (g) At what rate does this force do work on the rod?

Short Answer

Expert verified

The magnitude of induced emf is0.60 V.
The direction of induced emf is upside.
The value of current is 1.5 A.
The direction of current is clockwise.
The rate of thermal energy is 0.90 W.
Magnitude of applied force is, $F_{e x t} = 0.18 N$ .
Rate of external work done is 0.90 W.

Step by step solution

Step 1: Given

Length of conducting rod , $L = 10.0 c m = 10 脳 10^{- 2} m$
Uniform magnetic field coming out of page $\overset{鈫�}{B} = B_{0} \hat{k}$ , $B_{0} = 1.2 T$
Velocity of conducting rod $\overset{鈫�}{v} = - v_{0} \hat{i}$ , $v_{0} = 5.0 m / s$
Resistance of the rod is $R = 0.40 惟$

Determining the concept

Use Faraday鈥檚 law of electromagnetic induction with Lenz law. The rod is moving in a uniform magnetic field, so it experiences a force due to the applied magnetic field. Also, mechanical power is the dot product of force and velocity.

Faraday'slaw of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Lenz's law states that the current induced in a circuit due to a change in a magnetic field is directed to oppose the change in flux and to exert a mechanical force that opposes the motion

Formulae are as follows:

$蠒_{B} = 鈭� \overset{鈫�}{B} 路 d \overset{鈫�}{s}$

${\overset{,}{O}}_{i n d} = - \frac{d 蠒_{B}}{d t} \overset{鈫�}{F} = i \overset{鈫�}{d l} 脳 \overset{鈫�}{B} P = i^{2} R P = \overset{鈫�}{F} 路 \overset{鈫�}{v}$

Where, $桅$ is magnetic flux, B is magnetic field, i is current, R is resistance,饾渶 is emf, l islength, F is force, v is velocity, P is power

(a) Determining the magnitude of induced emf

$d 蠒_{B} = \overset{鈫�}{B} 路 d \overset{鈫�}{A} = B_{0} \hat{k} (- d A) \hat{k} = - B_{0} d A$

Where, $d A = d x 路 L$

${\overset{,}{O}}_{i n d} = - \frac{d 蠒_{B}}{d t} {\overset{,}{O}}_{i n d} = - \frac{d}{d t} (- B_{0} d A) {\overset{,}{O}}_{i n d} = - \frac{d}{d t} B_{0} d x 路 L {\overset{,}{O}}_{i n d} = B_{0} 路 L 路 \frac{d x}{d t} {\overset{,}{O}}_{i n d} = B_{0} L v_{0} {\overset{,}{O}}_{i n d} = B_{0} L v = 1.2 脳 10 脳 10^{- 2} 脳 5.0 {\overset{,}{O}}_{i n d} = 0.60 V$

Hence, the magnitude of induced emf is0.60 V.

(b) Determining the direction of induced emf

By Lenz law, as the applied magnetic field is along z-axis, to oppose that, the induced magnetic field will be along negative z 鈥� axis. Therefore, the emf must be directed up the page.

Hence, the direction of induced emf is upside.

(c) Determining the Value of current

By using Ohms law, findthe current throughthe rod.

$i_{i n d} = \frac{{\overset{,}{O}}_{i n d}}{R} i_{i n d} = \frac{0.60}{0.40} V i_{i n d} = 1.5 A$

Hence, the value of current is 1.5 A.

(d) Determining the direction of current in the conducting loop

Direction ofthe current is determined by the direction of emf, so it is clockwise.

Hence, the direction of current is clockwise.

(e) Determining the rate of thermal energy generated in the rod

The rate at which heat is generated in the rod is electrical power, so ,

$P = i_{i n d}^{2} R P = 1.5 脳 1.5 脳 0.40 P = 0.90 W$

Hence, the rate of thermal energy is 0.90 W.

(f) Determining the value of external force to maintain v→t

The force acting onthe current loop kept in magnetic field is, $\overset{鈫�}{F} = i \overset{鈫�}{d l} 脳 \overset{鈫�}{B}$

$\overset{鈫�}{F} = i \overset{鈫�}{d l} 脳 \overset{鈫�}{B} = i_{i n d} L \hat{j} 脳 B_{0} \hat{k} = i_{i n d} L B_{0} \hat{i}$

So thevalue of external force to maintain

${\overset{鈫�}{F}}_{e x t} = i_{i n d} L B_{0} \hat{i} = - 1.5 脳 10 脳 10^{- 2} 脳 1.2 \hat{i} = - 0.18 \hat{i} N$

Hence, magnitude of applied force is, role="math" localid="1661918931404" $F_{e x t} = - 0.18 N$ .

(g) Determining the rate of external work

Rate of external work:

Mechanical power

$P = \overset{鈫�}{F_{e x t}} 路 \overset{鈫�}{v} P = (- 0.18 \hat{i}) 路 (- v_{0} \hat{i}) P = 0.18 脳 5.0 P = 0.90 W$

Hence, rate of external work done is 0.90 W.

Therefore, the rate of external work is nothing but the power delivered to the specific process. In this case, to move the rod, power is spent by the external force.