The magnetic force per meter on a wire is measured to be only 45 %% of its maximum possible value. Calculate the angle between the wire and the magnetic field.

Answers

Answer 1

Answer:

27°

Explanation:

The force is proportional to the sine of the angle between the wire and the magnetic field. (See the ref.)

So theta = arcsin(0.45)

=27°

Answer 2

The angle between the wire and the magnetic field is 27°.

Calculation of the angle:

Since The magnetic force per meter on a wire is measured to be only 45 %

So here we know that The force should be proportional to the sine of the angle between the wire and the magnetic field

Therefore,

theta = arcsin(0.45)

=27°

Hence, The angle between the wire and the magnetic field is 27°.

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Related Questions

Suppose you are looking into the end of a long cylindrical tube in which there is a uniform magnetic field pointing away from you. If the magnitude of the field is decreasing with time the direction of the induced magnetic field is

Answers

Answer:

If the magnitude of the field is decreasing with time the direction of the induced magnetic field is CLOCKWISE

Explanation

This is because If the magnetic field decreases with time, the electric field will be produced in order to oppose the change in line with lenz law. Thus The right hand rule can be applied to find that the direction of electric field is in the clockwise direction.

The 2-Mg truck is traveling at 15 m/s when the brakes on all its wheels are applied, causing it to skid for 10 m before coming to rest. The total mass of the boat and trailer is 1 Mg. Determine the constant horizontal force developed in the coupling C, and the friction force developed between the tires of the truck and the road during this time.

Answers

Answer:

constant horizontal force developed in the coupling C = 11.25KN

the friction force developed between the tires of the truck and the road during this time is 33.75KN

Explanation:

See attached file

The friction force between the tires of the truck and the road is 22500 N.

Calculating the friction force:

It is given that a 2 Mg truck ( m = 2000 Kg) is initially moving with a speed of u = 15 m/s.

Distance traveled before coming to rest, s = 10m

The final velocity of the truck will be zero, v = 0

When the breaks are applied, only the frictional force is acting on the truck and it is opposite to the motion of the truck.

The frictional force is given by:

f = -ma

the acceleration of the truck = -a

The negative sign indicates that the acceleration is opposite to the motion.

Applying the third equation of motion we get:

v² = u² -2as

0 = 15² - 2×a×10

225 = 20a

a = 11.25 m/s²

So the magnitude of frictional force is:

f = ma = 2000 × 11.25 N

f = 22500 N

Learn more about friction force:

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