Question

One proton enters in a magnetic field of $2500 \,\text{N/Amp}\cdot\text{m}$ intensity with velocity of $4\times10^5\,\text{m/s}$ in parallel to the field. The force exerted on proton will be

• A. $0\,\text{N}$
• B. $4.8\times10^{-10}\,\text{N}$
• C. $0.48\times10^{-10}\,\text{N}$
• D. $4.8\times10^{10}\,\text{N}$

Hint:

Magnetic force is maximum when the particle moves perpendicular to the field and zero when it moves parallel to the field.

Answer 1

The Correct Option is A

Concept: When a charged particle moves in a magnetic field, the magnetic force acting on it is given by \[ F=qvB\sin\theta \] where \[ q=\text{charge of particle} \] \[ v=\text{velocity of particle} \] \[ B=\text{magnetic field intensity} \] \[ \theta=\text{angle between velocity and magnetic field} \] The magnetic force depends on the sine of the angle between the velocity vector and magnetic field vector.

Step 1: Identify the given quantities. \[ B=2500\,\text{N/Amp}\cdot\text{m} \] \[ v=4\times10^5\,\text{m/s} \] The proton is moving

parallel to the magnetic field. Therefore, \[ \theta=0^\circ \]

Step 2: Apply the magnetic force formula. \[ F=qvB\sin\theta \] Substituting \(\theta=0^\circ\), \[ F=qvB\sin0^\circ \] Since \[ \sin0^\circ=0 \] we get \[ F=0 \]

Step 3: Interpret the result. Whenever a charged particle moves parallel or antiparallel to a magnetic field, no magnetic force acts on it because the velocity and field vectors are along the same line. center minipage0.35

Force on proton = $0\,\text{N}$ minipage center