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CBSE Class 12 Physics 2023 Delhi Set 3 Paper

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Question : 11 of 14
Marks: +1, -0
(a) Two identical circular loops PP and QQ, each of radius RR carrying current II are kept in perpendicular planes such that they have bos\text{bo} s common centre OO as shown in the figure.
Find the magnitude and direction of the net magnetic field at point OO.
OR
(b) A long straight conductor kept along X′XX' X axis, caries a steady current II along +x+x direction. At an instant tt, a particle of mass mm and charge qq at point (x,y)(x, y) moves with a velocity v→\overset{\rightarrow}{v} along +y+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 PP
BP→=μ0nI2Rj^\overset{\rightarrow}{B_P} = \frac{\mu_0 n I}{2 R} \hat{j}
Magnetic field due to loop QQ
BQ→=μ0nI2Ri^\overset{\rightarrow}{B_Q} = \frac{\mu_0 n I}{2 R} \hat{i}
BNET→=μ0nI2Ri^+μ0nI2Rj^\overset{\rightarrow}{B_{NET}} = \frac{\mu_0 n I}{2 R} \hat{i} + \frac{\mu_0 n I}{2 R} \hat{j}
=μ0nI2R(i^+j^)= \frac{\mu_0 n I}{2 R} (\hat{i}+\hat{j})
Its magnitude=∣BNET→∣\text{Its magnitude} = \left| \overset{\rightarrow}{B_{NET}} \right|
=μ0nI2R12+12= \frac{\mu_0 n I}{2 R} \sqrt{1^2+1^2}
=μ0nI2R= \frac{\mu_0 n I}{\sqrt{2} R}
Its direction,
B^NET=BNET→∣BNET→∣\hat{B}_{NET} = \frac{ \overset{\rightarrow}{B_{NET}} }{ \left| \overset{\rightarrow}{B_{NET}} \right| }
=i^+j^2= \frac{ \hat{i} + \hat{j} }{ \sqrt{2} }
So, the direction will be 45∘45^{\circ} with B→P\overset{\rightarrow_P}{B} and B→Q\overset{\rightarrow_Q}{B}.
(b) The magnetic due to the straight conductor is μ0I2πyk^\frac{\mu_0 I}{2 \pi y} \hat{k}
Velocity of the charged particle is vj^v \hat{j}
So, the force acting on the particle is
F→=q(v→×B→)\overset{\rightarrow}{F} = q \left(\overset{\rightarrow}{v} \times \overset{\rightarrow}{B}\right)
or, F→=q[vj^×μ0I2πyk^]\overset{\rightarrow}{F} = q \left[ v \hat{j} \times \frac{\mu_0 I}{2 \pi y} \hat{k} \right]
or, F→=qvμ0I2πy[j^×k^]\overset{\rightarrow}{F} = \frac{q v \mu_0 I}{2 \pi y} [ \hat{j} \times \hat{k} ]
∴ F→=qvμ0I2πyj^\overset{\rightarrow}{F} = \frac{q v \mu_0 I}{2 \pi y} \hat{j}
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