The magnitude of maximum magnetic force that could be exerted on the proton is 1.44 x 10^-12 N.
The magnitude of the force on a charged particle moving in a magnetic field is given by the formula,
F= qvB
Here q is the charge on proton = 1.6 x 10^-19 C.
v is the velocity with which the particle is moving = 6.00 x 10^6 m/s
And B is the value of the magnetic field = 1.5 T
Putting the given values in the above equation,
F = 1.6 x 10^-19 x 6 x 10^6 x 1.5 = 1.44 x 10^-12 N.
Hence, the magnitude of maximum magnetic force that could be exerted on the proton is 1.44 x 10^-12 N.
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The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.
Answer:
Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W
Explanation:
As we know that the current in the circuit at given instant of time is
i = 2.0 mA
R = 10 k ohm
now we know by ohm's law
so voltage across the capacitor + voltage across resistor = V
Now we know that
here rate of change in energy of the capacitor is given as
