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4 years ago

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A conducting coil of 1785 turns is connected to a galvanometer, and the total resistance of the circuit is 43.9 Ω. The area of e

ach turn is 4.50 10-4 m2. This coil is moved from a region where the magnetic field is zero into a region where it is nonzero, the normal to the coil being kept parallel to the magnetic field. The amount of charge that is induced to flow around the circuit is measured to be 9.33 10-3 C. Find the magnitude of the magnetic field.

1 answer:

34kurt4 years ago

6 0

Answer:

0.5099 T

Explanation:

We are given that

Number of turns=N=1785

Resistance of circuit, R= $43.9\Omega$

Area of each turn, $A=4.5\times 10^{-4} m^2$

Charge , q= $9.33\times 10^{-3} C$

We have to find the magnitude of the magnetic field.

We know that magnetic field, $B=\frac{qR}{NA}$

Substitute the values

Magnetic field, B= $\frac{9.33\times 10^{-3}\times 43.9}{1785\times 4.5\times 10^{-4}}=0.5099 T$

Hence, the magnitude of the magnetic field=0.5099 T

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A 4.00 kg object is moving at 5.00 m/s NORTH. It strikes a 6.00 kg object that is moving WEST at 2.00 m/s. The objects undergo a

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$\Delta K=K_f-K_0$

The final kinetic energy is:

$K_f=\frac{1}{2}(m_1+m_2)v^2_f$

We use the sum of the masses because the objects are stuck together after the collision. The initial kinetic energy is:

$K_0=\frac{1}{2}m_1v^2_{01}+\frac{1}{2}m_2v^2_{02}$

Substituting we get:

$\Delta K=\frac{1}{2}(m_1+m_2)v^2_f-(\frac{1}{2}m_1v^2_{01}+\frac{1}{2}m_2v^2_{02})$

The only missing variable is the final velocity. To determine the final velocity we will use the conservation of momentum.

We will use the conservation of momentum in the horizontal direction (west) and the conservation of momentum in the vertical direction (north).

In the horizontal direction we have:

$m_1v_{h1}+m_2v_{h2}=(m_1+m_2)v_{hf}$

Since the object 1 has no velocity in the horizontal direction we have that:

$\begin{gathered} m_1(0)+m_2v_{h2}=(m_1+m_2)v_{hf} \\ m_2v_{h2}=(m_1+m_2)v_{hf} \end{gathered}$

Now we solve for the final horizontal velocity:

$\frac{m_2v_{h2}}{\mleft(m_1+m_2\mright)}=v_{hf}$

Now we substitute the values:

$\frac{(6kg)(2\frac{m}{s})}{(4kg+6kg)}=v_{hf}$

Solving the operations we get:

$1.2\frac{m}{s}=v_{hf}$

Now we use the conservation of momentum in the vertical direction, we get:

$m_1v_{v1}+m_2v_{v2}=(m_1+m_2)v_{vf}$

Since the second object has no vertical velocity we get:

$m_1v_{v1}=(m_1+m_2)v_{vf}$

Now w solve for the final vertical velocity, we get:

$\frac{m_1v_{v1}}{\mleft(m_1+m_2\mright)}=v_{vf}$

Now we substitute the values:

$\frac{(4kg)(5\frac{m}{s})}{(4kg+6kg)}=v_{vf}$

Now we solve the operations:

$2\frac{m}{s}=v_{vf}$

Now we determine the magnitude of the final velocity using the following formula:

$v_f=\sqrt[]{v^2_{hf}+v^2_{vf}}$

Substituting the values:

$v_f=\sqrt[]{(1.2\frac{m}{s})^2+(2\frac{m}{s})^2}$

Solving the operations:

$v_f=2.53\frac{m}{s}$

Now we substitute this in the formula for the kinetic energy and we get:

$\Delta K=\frac{1}{2}(4kg+6kg)(2.53\frac{m}{s})^2-(\frac{1}{2}(4kg)(5\frac{m}{s})^2+\frac{1}{2}(6kg)(2\frac{m}{s})^2)$

Solving the operations:

$\Delta K=32J-62J=-30J$

Therefore, there was a loss of 30J of kinetic energy.

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1 year ago