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scZoUnD [109]
3 years ago
5

When is the force on a current-carrying wire in a magnetic field at its strongest?

Physics
2 answers:
statuscvo [17]3 years ago
7 0

Answer: The correct answer is 'when the current is at a 90-degree angle to the field'

Explanation:

The Force on a current carrying wire in a magnetic field is given by lorents Force:

F=BILsin\theta

B= Magnetic Field in Tesla

I = Current in a wire in Ampere

L= Length of the wire in meters

\theta = angle between the Vetor B and vector I in degrees.

F=BILsin90^o (force will be maximum)

F=BILsin0^o (force will be minimum)

3241004551 [841]3 years ago
3 0
Its D <span>-when the current is at a 90-degree angle to the field</span>
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ASAP
scoundrel [369]

Answer:

A

Explanation:

Hooke's law! F(spring)=-kx

There's no tricky square law here. The spring constant doesn't change, only x (distance stretched) changes. Therefore, if distance is halved, Force will be halved.

5 0
3 years ago
A meter stick is suspended vertically at a pivot point 22 cm from the top end. It is rotated on the pivot until it is horizontal
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Answer:

5.82812 rad/s

Explanation:

L = Length of meter stick = 1 m = 100 cm

m_c = The center of mass of the stick = \frac{L}{2}-0.22=0.5-0.22=0.28\ m

\omega = Angular velocity

Moment of inertia of the system is given by

I=I_c+mr^2\\\Rightarrow I=\frac{mL^2}{12}+mr^2\\\Rightarrow I=\frac{m1^2}{12}+m0.28^2\\\Rightarrow I=m(\frac{1}{12}+0.0784)

As the energy in the system is conserved

mgh=I\frac{\omega^2}{2}\\\Rightarrow mgh=m(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow gh=(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow \omega=\sqrt{\frac{2gh}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=\sqrt{\frac{2\times 9.81\times 0.28}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=5.82812\ rad/s

The maximum angular velocity is 5.82812 rad/s

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