Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 20, Problem 3P

(a)

To determine

The current in the metal rod.

(a)

Expert Solution
Check Mark

Answer to Problem 3P

The current in the metal rod is 3.3×105A_.

Explanation of Solution

Physics, Chapter 20, Problem 3P

Figure 1

Write the expression for the induced emf in the metal rod.

ε=vBL        (I)

Here, ε is the induced emf, L is the length of the rod, v is the velocity.

Write the expression for the current in the metal rod.

I=εR        (II)

Here, I is the current, R is the resistance.

Use equation (II) in (I) to solve for I.

I=vBLR        (III)

Conclusion:

Substitute 16cm/s for v, 0.75T for B, 5.0cm for L, 180Ω for R in equation (III) to find I.

    I=(16cm×102m1cm/s)(0.75T)(5.0cm×102m1cm)(180Ω)=0.0060V180Ω=3.3×105A

Therefore, the current in the metal rod is 3.3×105A_.

(b)

To determine

The energy dissipated in the resistor.

(b)

Expert Solution
Check Mark

Answer to Problem 3P

The rate of energy dissipated in the resistor is 2.0×107W_.

Explanation of Solution

Write the expression for the energy dissipated in the resistor.

P=I2R        (IV)

Here, P is the power or the rate of energy dissipated in the resistor.

Conclusion:

Substitute 3.3×105A for I, 180Ω for R in equation (IV) to find P.

    P=(3.3×105A)2(180Ω)=2.0×107W

Therefore, the energy dissipated in the resistor is 2.0×107W_.

(c)

To determine

The magnetic force on the metal rod.

(c)

Expert Solution
Check Mark

Answer to Problem 3P

The magnetic force on the metal rod is 1.3×106N in the left direction_.

Explanation of Solution

Write the expression for the magnetic force.

    F=IL×B        (V)

Here, F is the magnetic force.

Conclusion:

Substitute 3.3×105A for I, 5.0cmj^ for L, 0.75T(k^) for B in equation (V) to find F.

    F=(3.3×105A)(5.0cmj^)×(0.75T(k^))=(3.3×105A)(5.0cm)(0.75T)(j^×k^)=1.3×106N(i^)

Therefore, the magnetic force on the metal rod is 1.3×106N in the left direction_.

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Chapter 20 Solutions

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