Q. 385.0( 1 Vote )

# A particle of mass m and positive charge q, moving with a uniform velocity v, enters a magnetic field B as shown in figure.

(a) Find the radius of the circular arc it describes in the magnetic field.

(b) Find the angle subtended by the arc at the centre.

(c) How long does the particle stay inside the magnetic field?

(d) Solve the three parts on the above problem if the charge q on the particle is negative.

Answer :

Given-

Mass of the particle = *m*

Positive charge on the particle = *q*

velocity of the particle = *v*

Magnetic field = *B*

(a)The radius of the circular arc described by the particle in the magnetic field-

We know,

The radius of the circle is given by,

where,

*m* is the mass of a proton

v= velocity of the particle

B = magnetic force

q= charge on the particle = C

(b) The angle subtended by the arc at the centre

Line MAB is the tangent to arc ABC,

When the particle enters into the magnetic field, it follows a path in the form of arc as shown in fig.

Now, the angle described by the charged particle is nothing but the angle ∠ MAO is ,

∠MAO = 90°

Also from fig , ∠NAC = 90°

∠OAC = ∠ OCA = *θ*

Then, by angle-sum property of a triangle, sum of all angles of a triangle is 180^{0}

∠AOC =180°− (*θ* + *θ*)

=

(c)The time for which the particle stay inside the magnetic field is given by-

Distance covered by the particle inside the magnetic field,is the length of arc subtended by angle θ and the radius

*l* = *r*θ

from (1)

*l*= *r*

time taken for a complete cycle will be its circumference and

the velocity is V -

Also The radius of the circle is given by,

where,

*m* is the mass of a proton

v= velocity of the particle

B = magnetic force

q= charge on the particle = C

(d) If the charge *q* on the particle is negative, then

(i) Radius of circular arc is given by

(ii) The centre of the arc lies within the magnetic field

Therefore, the angle subtended by the arc =

(iii) The time taken by the particle to cover the path inside the magnetic field

Rate this question :

A rectangular coil of 100 turns has length 5 cm and width 4 cm. It is placed with its plane parallel to a uniform magnetic field and a current of 2A is sent through the coil. Find the magnitude of the magnetic field B, if the torque acting on the coil is 0.2 N m^{–1}.

A circular coil of radius 2.0 cm ahs 500 turns in it and carries a current of 1.0 A. Its axis makes an angle of 30° with the uniform magnetic field of magnitude 0.40 T that exist in the space. Find the torque acting on the coil.

HC Verma - Concepts of Physics Part 2A rectangular loop of sides 20 cm and 10 cm carries a current of 5.0 A. A uniform magnetic field of magnitude 0.20 T exists parallel to the longer side of the loop.

(a) What is the force acting on the loop?

(b) What is the torque acting on the loop?

HC Verma - Concepts of Physics Part 2

A circular loop carrying a current i has wire of total length L. A uniform magnetic field B exists parallel to the plane of the loop.

(a) Find the torque on the loop.

(b) If the same length of the wire is used to form a square loop, what would be the torque? Which is larger?

HC Verma - Concepts of Physics Part 2

Fe^{+}ions are acceleration through a potential difference of 500 V and are injected normally into a homogeneous magnetic field B of strength 20.0 mT. Find the radius of the circular paths followed by the isotopes with mass numbers 57 and 58. Take the mass of an ion = A(1.6 × 10^{–27}) kg where A is the mass number.

A charged particle moves along a circle under the action of possible constant electric and magnetic fields. Which of the following are possible?

HC Verma - Concepts of Physics Part 2A charged particle moves in a gravity-free space without change in velocity. Which of the following is/are possible?

HC Verma - Concepts of Physics Part 2If a charged particle at rest experiences no electromagnetic force,

HC Verma - Concepts of Physics Part 2If a charged particle goes unaccelerated in a region containing electric and magnetic fields.

HC Verma - Concepts of Physics Part 2