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DENIUS [597]
3 years ago
12

A wire that is 0.65 m long and carrying a current of 8.2 A is at right angles to a uniform magnetic field. The force on the wire

is 0.40 N. What is the strength of the magnetic field?
Physics
1 answer:
Luda [366]3 years ago
3 0

Answer:

0.075 T

Explanation:

When a current-carrying wire is immersed in a region with magnetic field, the wire experiences a force, given by

F=ILB sin \theta

where

I is the current in the wire

L is the length of the wire

B is the strength of the magnetic field

\theta is the angle between the direction of I and B

In this problem we have:

L = 0.65 m is the length of the wire

I = 8.2 A is the current in the wire

F = 0.40 N is the force experienced by the wire

\theta=90^{\circ} since the current is at right angle with the magnetic field

Solving the formula for B, we find the strength of the magnetic field:

B=\frac{F}{IL sin \theta}=\frac{0.40}{(8.2)(0.65)(sin 90^{\circ})}=0.075 T

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gas has a volume of 185 ml and pressure of 310 mm hg. The desiered volume is 74.0 ml. What is the required new pressure
Mamont248 [21]

Answer:

The required new pressure is 775 mm hg.

Explanation:

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Let the required new pressure be '\text{P}_2'.

As we know that Boyle's law formula states that;

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where, P_1 = original pressure of gas in the container = 310 mm hg

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So,  P_2 = \frac{P_1 \times V_1}{V_2}  

       P_2 = \frac{310 \times 185}{74}

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Hence, the required new pressure is 775 mm hg.

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