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Oliga [24]
3 years ago
7

A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

and resistivity of 2.18 × 10-8Ω·m. At what rate must the magnitude of the magnetic field change to induce a 11 A current in the loop?
Physics
1 answer:
I am Lyosha [343]3 years ago
8 0

Answer:

Rate of change of magnetic field is 3.466\times 10^3T/sec        

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop r=\frac{0.13}{2}=0.065m

Length of the circular loop L=2\pi r=2\times 3.14\times 0.065=0.4082m

Wire is made up of diameter of 2.6 mm

So radius r=\frac{2.6}{2}=1.3mm=0.0013m

Cross sectional area of wire A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2

Resistivity of wire \rho =2.18\times 10^{-8}m

Resistance of wire R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm

Current is given i = 11 A

So emf  e=11\times 1.67\times 10^{-3}=0.0183volt

Emf induced in the coil is e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}

0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}

\frac{dB}{dt}=3.466\times 10^3=T/sec

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Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f
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Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

W_T+W_F=K_f-K_i (1)

Where:

W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to \frac{1}{2}mv^{2}, so if the initial velocity v_i is equal to zero, the initial kinetic energy K_i is equal to zero.

Then, replacing the values on the equation and solving for v_f, we get:

W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}

\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f

So, the speed after being pulled 3.2m is 2.4123 m/s

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Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea
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Answer:

6.5 m/s

Explanation:

We are given that

Distance, s=100 m

Initial speed, u=1.4 m/s

Acceleration, a=0.20 m/s^2

We have to find the final velocity at the end of the 100.0 m.

We know that

v^2-u^2=2as

Using the formula

v^2-(1.4)^2=2\times 0.20\times 100

v^2-1.96=40

v^2=40+1.96

v^2=41.96

v=\sqrt{41.96}

v=6.5 m/s

Hence, her final velocity at the end of the 100.0 m=6.5 m/s

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Answer:

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