CBSE Class 12 Physics 2023 Delhi Set 3 Paper

Question : 11 of 14

Marks: +1, -0

(a) Two identical circular loops

P

and

Q

, each of radius

R

carrying current

I

are kept in perpendicular planes such that they have

\bo {s}

common centre

{\O}

as shown in the figure.

Find the magnitude and direction of the net magnetic field at point

O

.
OR
(b) A long straight conductor kept along

{\X}^{'} {\X}

axis, caries a steady current

{\I}

along

+x

direction. At an instant

t

, a particle of mass

m

and charge

q

at point

(x, y)

moves with a velocity

{v}↖{→}

along

+y

direction. Find the magnitude and direction of the force on the particle due to the conductor.

Solution:

(a)

Magnetic field due to loop

{\P}

{B_P}↖{→}=\;{µ_0 n I}/{2 R} {j}↖{\^}

Magnetic field due to loop

{\Q}

{B_Q}↖{→}\;=\;{µ_0 n I}/{2 R} {i}↖{\^}

{B_{N E T}}↖{→}\;=\;{µ_0 n I}/{2 R} {i}↖{\^}+\;{µ_0 n I}/{2 R} {j}↖{\^}

\;=\;{µ_0 n I}/{2 R}({i}↖{\^}+{j}↖{\^})

\;\text " Its magnitude "\;\;={| {B_{N E T}}↖{→} |}

\;=\;{µ_0 n I}/{2 R} {√{1^2+1^2}}

\;=\;{µ_0 n I}/{{√{2}} R}

Its direction,

{B_{N E T}↖{\^}}=\;{{B_{N E T}}↖{→}}/{{| {B_{N E T}}↖{→} |}}

=\;{\;{µ_0 n I}/{2 R}({i}↖{\^}+{j}↖{\^})}/{{µ_0 n I}/{{√{2}} R}}=\;{{i}↖{\^}+{j}↖{\^}}/{{√{2}}}

So, the direction will be

45^{∘}

with

{B}↖{→}_P

and

{B}↖{→}_Q

.
(b) The magnetic due to the straight conductor is

\;{µ_0 I}/{2 π y} {k}↖{\^}

Velocity of the charged particle is

v {j}↖{\^}

So, the force acting on the particle is

\;{F}↖{→}=q({v}↖{→} × {B}↖{→})

or,

\;{F}↖{→}=q [v {j}↖{\^} × \;{µ_0 I}/{2 π y} {k}↖{\^}]

or,

\;{F}↖{→}=\;{q v µ_0 I}/{2 π y}[{j}↖{\^} × {k}↖{\^}]

∴

\;{F}↖{→}=\;{q v µ_0 I}/{2 π y} {j}↖{\^}

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