Answer:
[tex]F=9.09\times 10^{-7}\ N[/tex]
Explanation:
Given that,
Charge, q = 0.250 μC
It is released from a height of 30 cm or 0.03 m
The magnetic field strength is 1.50 T.
First we find the velocity using the conservation of energy as follows :
[tex]mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 0.3} \\\\v=2.424\ m/s[/tex]
Now, the magnetic force is given by :
[tex]F=qvB\\\\=0.25\times 10^{-6}\times 2.424\times 1.5\\\\=9.09\times 10^{-7}\ N[/tex]
So, the magnetic force is [tex]9.09\times 10^{-7}\ N[/tex]. Since, the bob is at the lowest point, the direction of the magnetic force at the lowest point is upward.
