Solveeit Logo
Login

Question

Question: A proton enters a magnetic field of flux density \(1.5Wb/{m^2}\) with a speed of \(2 \times {10^7}m/...

A proton enters a magnetic field of flux density 1.5Wb/m21.5Wb/{m^2} with a speed of 2×107m/s2 \times {10^7}m/s at an angle of 30∘30^\circ with the field. The force on a proton will be
1.0.44×10−12N0.44 \times {10^{ - 12}}N
2.2.4×10−12N2.4 \times {10^{ - 12}}N
3.24×10−12N24 \times {10^{ - 12}}N
4.0.024×10−12N0.024 \times {10^{ - 12}}N

Explanation

Solution

Here a proton is entering a magnetic field H. The proton will feel a force on it due to the magnetic field. The force on the proton will be proportional to the magnetic flux density and the velocity of the proton. Here there is a formula for calculating the force on a proton due to a magnetic field. Apply the formula and solve.

Formula Used:
The formula for the force on the proton is given by
F=qvBsin⁡θF = qvB\sin \theta
Where;
F = Force;
q = charge on proton.
v = velocity or speed;
B = magnetic flux density.

Complete answer:
Find out the force on a proton which is entering in a magnetic field. With magnetic flux B = 1.5Wb/m21.5Wb/{m^2} and velocity v =2×107m/s2 \times {10^7}m/s. A positively charged particle which is put in a magnetic field will feel a force due to the presence of a magnetic field if the positive charge is moving relative to this field. Together the combination of these two effects creates a force that we call the Lorentz force. A positively charged particle which is moving through a magnetic field with a strength B and with a velocity v will feel the Lorentz force: F=qvBsin⁡θF = qvB\sin \theta . The magnetic force is zero for motion parallel to the field; this produces a spiral motion than a circular one
The formula for force is:
F=qvBsin⁡θF = qvB\sin \theta
Put the known values in the above equation; Take q = 1.6×10−191.6 \times {10^{ - 19}}.
F=(1.6×10−19)×1.5×2×107×sin⁡30F = (1.6 \times {10^{ - 19}}) \times 1.5 \times 2 \times {10^7} \times \sin 30
Do the needed calculation:
F=3.2×1.5×10−12sin⁡30F = 3.2 \times 1.5 \times {10^{ - 12}}\sin 30
Here, sin30 = 0.5
F=3.2×10−12×0.5×1.5F = 3.2 \times {10^{ - 12}} \times 0.5 \times 1.5
The value of force is:
F=2.4×10−12NF = 2.4 \times {10^{ - 12}}N

Option “3” is correct. The force on a proton will be 2.4×10−12N2.4 \times {10^{ - 12}}N.

Note:
Here we need one equation for force on a proton in a magnetic field and rest of the values are already given, just put in the given values in the equation and solve for the unknown and also make sure to take the value of proton to be 1.6×10−19C1.6 \times {10^{ - 19}}C.