Question

Seema’s uncle was advised by his doctor to have an MRI(Magnetic Resonance Imaging) scan of his brain. Her uncle felt it to be expensive and wanted to postpone it.
When Seema learnt about this, she took the help of her family and also approached the doctor, who also offered a substantial discount. She then convinced her uncle to undergo the test to enable the doctor to know the condition of his brain . The information thus obtained greatly helped the doctor to treat him properly. Based on the above paragraph, answer the question.
Assuming that MRI test was performed using a magnetic field of 0.1T , find the minimum and maximum values of the force that the magnetic field could exert on a proton moving with a speed of ${10^4}m/s$.Given charge of proton=$1.6 \times {10^{ - 19}}C$

Answer 1

The paragraph given above has no purpose except making the question longer, to solve this question you need to apply the concept of magnetic force on a moving charge (which is proton in this case) on a given magnetic force.

Complete step by step solution:
Here magnetic field, charge on the particle and speed of the particle is given
i.e; magnetic field $B = 0.1T$,
charge on particle $q = 1.6 \times {10^{ - 19}}$ and
speed of the particle $v = {10^4}m/s$
we know that force on a moving charged particle in a magnetic field is cross product of speed and magnetic field multiplied by charge,
i.e; $F = q(v \times B)$
$\Rightarrow F = qvB\sin \theta$

As we can see when the value of $\sin \theta$ is maximum then force will be maximum and when value of sin$\theta$is minimum then force will be minimum.
So force will be maximum when $\sin \theta = 1$or $\theta = 90$and
Force will be minimum when $\sin \theta = 0$or $\theta = 0$
Therefore, maximum and minimum force will be
${F_{{\rm{maximum}}}} = qvB \times 1 \\\ \Rightarrow{F_{{\rm{maximum}}}} = 1.6 \times {10^{ - 19}} \times {10^4} \times 0.1 \\\ \Rightarrow{F_{{\rm{maximum}}}} = 1.6 \times {10^{ - 16}}\,N$
This will be the case when the proton moves perpendicular to the direction of the magnetic field.
${F_{{\rm{minimum}}}} = qvB \times 0\\\ \therefore{F_{{\rm{minimum}}}} = 0\,N$
This will be the case when the proton moves parallel or antiparallel to the direction of the magnetic field.

Additional information:
A magnetic field is a vector field that determines the magnetic effect on electrical charges, electrical currents, and magnetised objects that pass. A force perpendicular to its own velocity and to the magnetic field is encountered by a charge travelling in a magnetic field.

Note: $F = q(v \times B)$, where $\theta$ is the angle between the directions of $v$ and $B$. This is the Lorentz force. In truth, in terms of the force on a charged particle travelling in a magnetic field, this is how we describe the magnetic field strength $B$.