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CBSE Class 12 Physics 2023 Delhi Set 3 Paper

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Question : 11
Total: 14
(a) Two identical circular loops P and Q, each of radius R carrying current I are kept in perpendicular planes such that they have 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:     👈: Video Solution
(a)

Magnetic field due to loop P
→
BP
=‌
µ0nI
2R
^
j

Magnetic field due to loop Q
→
BQ
‌=‌
µ0nI
2R
^
i

→
BNET
‌=‌
µ0nI
2R
^
i
+‌
µ0nI
2R
^
j

‌=‌
µ0nI
2R
(
^
i
+
^
j
)
‌ Its magnitude ‌‌=|
→
BNET
|
‌=‌
µ0nI
2R
√12+12
‌=‌
µ0nI
√2R

Its direction,
^
BNET
=‌
→
BNET
|
→
BNET
|
=‌
‌
µ0nI
2R
(
^
i
+
^
j
)
µ0nI
√2R
=‌
^
i
+
^
j
√2
So, the direction will be 45∘ with
→
B
P and
→
B
Q.
(b) The magnetic due to the straight conductor is ‌
µ0I
2πy
^
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
×‌
µ0I
2πy
^
k
]
or, ‌
→
F
=‌
qvµ0I
2πy
[
^
j
×
^
k
]
∴ ‌
→
F
=‌
qvµ0I
2πy
^
j
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