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A conducting coil of 1850 turns is connected to a galvanometer, and the total resistance of the circuit is 30 Ω. The area of each turn is 4.00 × 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 8.0 × 10-3 C. Find the magnitude of the magnetic field.

Respuesta :

Answer:

0.324 T

Explanation:

Parameters given:

Number of turns, N = 1850

Resistance, R = 30Ω

Area of each turn, A = [tex]4.00 * 10^{-4} m^2[/tex]

Charge in the circuit, q = [tex]8.0 * 10^{-3} C[/tex]

The induced EMF in the coil is given as:

[tex]V = \frac{-NBA}{t}[/tex]

EMF is also given in terms of current, I, and resistance, R, as:

V = IR = [tex]\frac{-NBA}{t}[/tex]

=> [tex]I*R*t = -NBA[/tex]

Charge, q, is the product of current and time. Hence:

It = q

=> [tex]q*R = -NBA[/tex]

Hence, magnetic field, B, will be:

[tex]B = \frac{qR}{-NA}[/tex]

[tex]B = \frac{8 * 10^{-3} * 30}{-1850 * 4 * 10^{-4}} \\\\\\B = -0.324 T[/tex]

The magnitude of magnetic field, |B| will be |-0.324| = 0.324 T

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