A particle of charge q and mass m is moving with a velocity ​ -vi(v≠0) ​ towards a large screen placed in the Y - Z plane at a distance d.

Question

A particle of charge q and mass m is moving with a velocity -v\(\hat i \)(v≠0) towards a large screen placed in the Y - Z plane at a distance d. If there is a magnetic field \(\vec B = B_0 \hat k\), the minimum value of v for which the particle will not hit the screen is

Answer 1

Correct answer is 

It should be maximum instead of minimum.

Answer 2

Correct option is (4) \(\frac{qdB_0}m\)

Step 1: Given data.

Mass of the particle = m

Particle of charge = q

Velocity = \(v\hat i\)

The magnetic field strength of the particle \(B = B_0\hat k\)

Step 2: Finding the minimum velocity v_min for which the particle will not hit the screen.

As we know in case of force on a charged particle in a magnetic field, the cyclotron equation is given as:

\(R = \frac{mv}{qB_0}\)   .......(i)

Where R is the radius of the circular path and v is the velocity.

Since, if the charged particle should not hit the screen then the radius of the circular path is less than distance d such as.

\(R \le d \left[\therefore R = \frac{mv}{qB_0}\right]\)

⇒ \(\frac{mv}{qB_0} \le d\)  [From equation (i)]

⇒ \(v \le \frac {qdB_0}{m}\)

Thus, the maximum value of  \(v = \frac {qdB_0}{m}\).