Chapter 24, Problem 94A

a.

To determine

The direction and magnitude of the force on the current carrying wire placed in the magnetic field, when the key is open.

Answer to Problem 94A

The force on the wire is $F=0\text{ N}$ .

Explanation of Solution

Given:

The key is open.

Formula Used:

  $F=\frac{V}{R}LB$

F is the force.

V is the voltage

R is the resistance

L is the length of the wire in the magnetic field

B is the magnetic field.

Calculations:

As the key is open, no current flows through the wire. Hence, no force is experienced by the wire as the wire does not produce magnetic field neither copper is a magnetic material.

  $F=0\text{ N}$

Conclusion:

The force on the wire is $F=0\text{ N}$ .

b.

To determine

The direction and magnitude of the force on the current carrying wire placed in the magnetic field, when the key is closed.

Answer to Problem 94A

The force on the wire is upwards of magnitude $F=0.62\text{ N}$ .

Explanation of Solution

Given:

The key is closed.

Voltage, $V=24\text{ V}$

Resistance, $R=5.5\Omega$

Magnetic field, $B=1.9\text{}\text{ T}$

The length of the wire in the magnetic field, $L=7.5\text{cm = 0}\text{.075}\text{m}$

Formula Used:

  $F=\frac{V}{R}LB$

F is the force.

V is the voltage

R is the resistance

L is the length of the wire in the magnetic field

B is the magnetic field.

Calculations:

  $F=\frac{V}{R}LB$

Substituting the values

  $F=\frac{24\text{ V}}{5.5\Omega }\left(0.075\text{ m}\right)\left(1.9\text{}\text{ T}\right)$

  $F=0.62\text{ N}$

The direction of the force can be determined by Fleming’s left-hand rule.Hence, the direction of force experienced by the wire is upwards.

Conclusion:

The wire experiences an upward force of $F=0.62\text{ N}$ .

c.

To determine

The direction and magnitude of the force on the current carrying wire placed in the magnetic field, when the key is closed and the direction of current is reversed.

Answer to Problem 94A

The force on the wire is downwards of magnitude $F=0.62\text{ N}$ .

Explanation of Solution

Given:

The key is closed.

Voltage $V=24\text{ V}$

Resistance $R=5.5\Omega$

Magnetic field $B=1.9\text{}\text{ T}$

The length of the wire in the magnetic field, $L=7.5\text{cm = 0}\text{.075}\text{m}$

Formula Used:

  $F=\frac{V}{R}LB$

F is the force.

V is the voltage

R is the resistance

L is the length of the wire in the magnetic field

B is the magnetic field.

Calculations:

  $F=\frac{V}{R}LB$

Substituting the values

  $F=\frac{\text{24 V}}{5.5\Omega }\left(0.075\text{ m}\right)\left(1.9\text{}\text{ T}\right)$

  $F=0.62\text{ N}$

The direction of the force can be determined by Fleming’s left-hand rule.Hence, the direction of force experienced by the wire is downwards.

Conclusion:

The wire experiences a downward force of $F=0.62\text{ N}$ .

d.

To determine

The direction and magnitude of the force on the current carrying wire placed in the magnetic field, when the key is closed and the wire has two $5.5\Omega$ resistors in series.

Answer to Problem 94A

The force on the wire is upwards of magnitude $F=0.31\text{ N}$ .

Explanation of Solution

Given:

The key is closed.

Voltage $V=24\text{ V}$

Total Resistance in series $R=5.5\Omega +5.5\Omega$

Magnetic field $B=1.9\text{}\text{ T}$

The length of the wire in the magnetic field, $L=7.5\text{cm = 0}\text{.075}\text{m}$

Formula Used:

  $F=\frac{V}{R}LB$

F is the force.

V is the voltage

R is the resistance

L is the length of the wire in the magnetic field

B is the magnetic field.

Calculations:

  $F=\frac{V}{R}LB$

Substituting the values

  $F=\frac{\text{24 V}}{5.5\Omega +5.5\Omega }\left(0.075\text{ m}\right)\left(1.9\text{}\text{ T}\right)$

  $F=0.31\text{ N}$

The direction of the force can be determined by Fleming’s left-hand rule.Hence, the force experienced by the wire is upward direction.

Conclusion:

The wire experiences an upwards force of $F=0.31\text{ N}$ .