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2 years ago

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One hundred turns of (insulated) copper wire are wrapped around a wooden cylindrical core of cross-sectional area 5.87 × 10-3 m2

. The two ends of the wire are connected to a resistor. The total resistance in the circuit is 14.9 Ω. If an externally applied uniform longitudinal magnetic field in the core changes from 1.93 T in one direction to 1.93 T in the opposite direction, how much charge flows through a point in the circuit during the change?

1 answer:

Grace [21]2 years ago

7 0

Answer:

The amount of charge that flow through a point is $\Delta q = 0.126 \ C$

Explanation:

From the question we are told that

The number of turns is $N = 100$

The area is $A = 5.87 *10^{-3} m^2$

The total resistance is $R = 14.9 \ \Omega$

The magnetic field in first direction is $B_1 = 1.93 \ T$

The magnetic field in second direction is $B_2 = -1.93 \ T$

The change in magnetic field is evaluated as

$\Delta B = B_1 - B_2$

substituting values

$\Delta B = 1.93 - [- 1.93]$

$\Delta B =3.2 \ T$

The induced emf due to the change is mathematically evaluated as

$e = NA \frac{\Delta B }{\Delta t }$

This can also be mathematically represented as

$e = IR$

Where I can be mathematically represented as

$I = \frac{\Delta q }{\Delta t}$

So

$e = \frac{\Delta q}{\Delta t } R$

Now

$\frac{\Delta q}{\Delta t } R = NA \frac{\Delta B }{\Delta t }$

=> $\Delta q = \frac{N A (\Delta B)}{R}$

substituting values

$\Delta q = \frac{100 * 5.87 *10^{-3} (3.2}{14.9}$

$\Delta q = 0.126 \ C$

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