Question

A conducting loop is made in the form of two squares of sides s₁ = 3cm and s₂ = 6.3 cm as shown. At time t = 0, the loop enters a region of length L = 17.3 cm that contains a uniform magnetic field B = 1.2 T, directed in the positive z-direction. The loop continues through the region with constant speed v = 35 cm/s. The resistance of the loop is R = 1.7 Ω.

1)At time t = t₁ = 0.029 s, what is I₁, the induced current in the loop? I₁ is defined to be positive if it is in the counterclockwise direction.

2)At time t = t₂ = 0.64 s, what is I₂, the induced current in the loop? I₂ is defined to be positive if it is in the counterclockwise direction.

3)What is F_x(t₂), the x-component of the force that must be applied to the loop to maintain its constant velocity v = 35 cm/s at t = t₂ = 0.64 s?

4)At time t = t₃ = 0.523 s, what is I₃, the induced current in the loop? I₃ is defined to be positive if it is in the counterclockwise direction.

5)Consider the two cases shown above. How does I_I, the magnitude of the induced current in Case I, compare to I_II, the magnitude of the induced current in Case II? Assume s₂ = 3s₁.

$h16_constantBD$

I_I < I_II

I_I = I_II

I_I > I_II

Answer 1

Given , Side of square 1 , s₁ = 3 cm = 0.03 m { 1 m = 100 cm }

Side of square 2 , s₂ = 6.3 cm = 0.063 m

Magnetic field , B = 1.2 T

Length of region of magnetic field , L = 17.3 cm = 0.173 m

Speed of loop = 35 cm/s = 0.35 m/s

Resistance of loop , R = 1.7 $\Omega$

( a ) The induced current , I₁ at time , t₁ = 0.029 second can be calculated as :

At t₁ = 0.029 s , distance of loop in magnetic field , d = v * t₁ = 0.035 * 0.029 = 0.0101 m = 1.01 cm

$\varepsilon = -\frac{d\phi }{dt}$

where $\varepsilon$ = Induced EMF or voltage

$\phi$ = flux

$\varepsilon_{1} = -\frac{d\phi }{dt} = B.v.s_{1}$

$\varepsilon_{1} = - 1.2*0.35*0.03$

$\varepsilon_{1} = - 0.0126 V$ { negative due to the increasing $\phi$ }

Now , $\varepsilon _{1}=I_{1}R$

$I_{1}=\frac{\varepsilon_{1} }{R}$

$I_{1}=\frac{-0.0126 }{1.7}=-0.0074 A=-7.4mA$

Hence , the induced current in the loop at t₁ = 0.029 second is - 0.0074 A or - 7.4 mA

( b ) The induced current , I₂ at time , t₂ = 0.64 second can be calculated as :

At t₂ = 0.64 s , distance of loop in magnetic field , d = v * t₂ = 0.35 * 0.64 = 0.224 m

distance of loop outside field = 0.224 - 0.173 = 0.051 m

{ This is greater than s₁, so we must be looking at an area of s₂ }

$\varepsilon_{2} = -\frac{d\phi }{dt} = B.v.s_{2}$

$\varepsilon_{2} = 1.2*0.35*0.063$

$\varepsilon_{2} = 0.02646V$    { positive due to the decreasing $\phi$ }

$\varepsilon _{2}=I_{2}R$

$I_{2}=\frac{\varepsilon_{2} }{R}$

$I_{2}=\frac{0.02646 }{1.7}=0.0156 A=15.6mA$

Hence , the induced current in the loop at t₂ = 0.64 second is 0.0156 A or 15.6 mA

( c ) The x-component of force , F_x at t₂ = 0.64 second can be calculated as :

Force , F is given by :

where I = current

L = length { L in this case is the length of the square in the magnetic field }

B = magnetic field

The bottom and top part of the square have forces that act in opposite directions and cancel .

The x - component of force is given by :

Hence , the x - component of force is 0.001179 Newton

( d ) The induced current , I₃ at time , t₃ = 0.523 second can be calculated as :

At t₃ = 0.523 s , distance of loop in magnetic field , d = v * t₃ = 0.35 * 0.523 = 0.183 m

distance of loop outside field  = 0.183 - 0.173 = 0.010 m

{ This is smaller than s₁, so we must be looking at an area of s₁ }

$\varepsilon_{3} = -\frac{d\phi }{dt} = B.v.s_{1}$

$\varepsilon_{3} = 1.2*0.35*0.03$

$\varepsilon_{3} = 0.0126V$ { positive due to the decreasing $\phi$ }

$\varepsilon _{3}=I_{3}R$

$I_{3}=\frac{\varepsilon_{3} }{R}$

$I_{3}=\frac{0.0126 }{1.7}=0.0074 A=7.4mA$

Hence , the induced current in the loop at t₃ = 0.523 second is -0.0074 A or 7.4 mA .