Chapter 20, Problem 30P

(a)

To determine

The constant speed $\left(v\right)$ should the bar be moved to produce a current $\left(I\right)$ of $1.00\text{A}$ in the resistor

Answer to Problem 30P

Solution: The constant speed $\left(v\right)$ should the bar be moved to produce a current $\left(I\right)$ of $1.00\text{A}$ in the resistor is $2.00\text{m}/\text{s}$

Explanation of Solution

Given Info: Resistance R is $6.00\text{Ω}$, length l is $1.20\text{m}$, current I is $1.00\text{A}$ and magnetic field B is $2.50\text{T}$,

The motional emf is

$\Delta V=Blv$

• $\Delta V$ is the motional emf,
• B is the magnetic field,
• l is the length of the bar,
• v is the speed of the bar,

Formula to calculate thecurrent is,

$I=\frac{\Delta V}{R}$

• I is the current in the rod,
• R is the resistance of the rod,

Use $Blv$ for $\Delta V$ in the above equation to find the expression for I

$I=\frac{Blv}{R}$

$v=\frac{IR}{Bl}$

Substitute $6.00\text{Ω}$ for R, $1.00\text{A}$ for I, $2.50\text{T}$ for Band $1.20\text{m}$ for lin the above equation to find $I$

$\begin{array}{c}I=\frac{\left(6.00\text{Ω}\right)\left(1.00\text{A}\right)}{\left(2.50\text{T}\right)\left(1.20\text{m}\right)}\\ =2.00\text{m}/\text{s}\end{array}$

Conclusion:

The constant speed $\left(v\right)$ should the bar be moved to produce a current $\left(I\right)$ of $1.00\text{A}$ in the resistor is $2.00\text{m}/\text{s}$

(b)

To determine

The power delivered to the resistor

Answer to Problem 30P

Solution: The power delivered to the resistor P is $6.00\text{W}$

Explanation of Solution

Given Info: Resistance R is $6.00\text{Ω}$, Current $I$ is $1.00A$.

Formula to calculate Power delivered to the resistor is,

$P=I^{2}R$

• P is the power delivered to the resistor,

Substitute $6.00\text{Ω}$ for $R$ and $1.00A$ for $I$ to find P.

$\begin{array}{c}P={\left(1.00\text{A}\right)}^2\left(\text{6}.00\text{Ω}\right)\\ =6.00\text{W}\end{array}$

Conclusion:

The power delivered to the resistor P is $6.00\text{W}$

(c)

To determine

The magnetic force $\left(F\right)$ is exerted on the moving bar.

Answer to Problem 30P

Solution: The magnetic force $\left(F\right)$ is exerted on the moving bar is $3.00\text{N}$.

Explanation of Solution

Given Info: Length l is $1.20\text{m}$, angle $\text{θ}$ is $90°$ current $I$ is $1.00\text{A}$, and magnetic field B is $2.50\text{T}$

Formula to calculate magnetic force $\left(F\right)$ is exerted on the moving bar.

$F=Blv\sin \theta$

• F is the magnetic force on the bar,

Substitute, $1.00\text{A}$ for I, $2.50\text{T}$ for B, $90°$ for $\theta$, and $1.20\text{m}$ for l in the above equation to find $I$.

$\begin{array}{c}F=\left(2.50\text{T}\right)\left(1.00\text{A}\right)\left(1.20\text{m}\right)\left(\sin 90°\right)\\ =3.00\text{N}\end{array}$

Conclusion:

The magnetic force $\left(F\right)$ is exerted on the moving bar is $3.00\text{N}$.

(d):

To determine

The power delivered by the force $F_{\text{app}}$ on the moving bar.

Answer to Problem 30P

Solution: The power delivered by the force $F_{\text{app}}$ on the moving bar is $6.00\text{W}$

Explanation of Solution

Given Info: speed vis $2.00\text{m}/\text{s}$ and $F_{\text{app}}$ is $3.00\text{N}$

The bar moves at the constant speed in the same direction. So, the acceleration is $a=0$

$F=F_{\text{app}}$

Formula to calculate power delivered by the force $F_{\text{app}}$ on the moving bar is

$P=F.v$

Use $F_{\text{app}}$ for F in the above equation to find the expression for P.

$P=F_{\text{app}}.v$

• P is the power delivered.
• $F_{\text{app}}$ is the force applied at instant

Substitute $3.00\text{N}$ for $F_{\text{app}}$ and $2.00\text{m}/\text{s}$ for vto find P

$\begin{array}{c}P=\left(3.00\text{N}\right)\left(2.00\text{m}/\text{s}\right)\\ =6.00\text{W}\end{array}$

Conclusion:

The power delivered by the force $F_{\text{app}}$ on the moving bar is $6.00\text{W}$