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Sindrei [870]
3 years ago
5

A 15-turn circular wire loop with a radius of 3.0 cm is initially in a uniform magnetic field with a strength of 0.5 T. The fiel

d decreases to zero over a time of 0.10 sec. What is the induced emf?a. 0.21 Vb. 7.1 Vc. 0.01 Vd. 0.2.25 V
Physics
1 answer:
lana66690 [7]3 years ago
6 0

To solve this problem it is necessary to apply the definition given in Faraday's law in a solenoid for which it is noted that

\epsilon = - d\frac{\phi_B}{dt}

\epsilon = -NA\frac{dB}{dt}

Where,

N = Number of loops

A = Cross sectional Area

B = Magnetic Field

\epsilon = (15)(\pi(0.03)^2)\frac{0-0.5}{0.1}

\epsilon = 0.212V

\epsilon = 0.21V

Therefore the correct answer is A.

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Follows are the solution to the given question:

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5 0
2 years ago
An electron moves with velocity v⃗ =(5.8i−6.7j)×104m/s in a magnetic field B⃗ =(−0.81i+0.60j)T.
Minchanka [31]

Answer:

Fₓ = 0,  F_{y} = 0  and  F_{z}<em> = - 3.115 10⁻¹⁵   N</em>

Explanation:

The magnetic force given by the expression

       F = q v xB

the bold are vectors,  the easiest analytical way to determine this force in solving the determinant

   F = q \left[\begin{array}{ccc}i&j&k\\5.8&-6.7&0\\-0.81&0.6&0\end{array}\right]  10^{4}

   F = 1.6 10⁻¹⁵ [ i( 0-0) + j (0-0) + k^( 5.8 0.60 - 0.81 67) ]

   F =i^0 + j^0   - k^  3.115 10⁻¹⁵   N

   

Fₓ = 0

F_{y} = 0

F_{z}<em> = - 3.115 10⁻¹⁵   N</em>

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