Two identical loops P and Q each of radius 5 cm are lying in
perpendicular planes such that they have a common centre as shown in
the figure. Find the magnitude and direction of the net magnetic field at
the common centre of the two coils, if they carry currents equal to A
and 4 A respectively.
Answers
Answer:
Radius of each coil R = 5 cm =0.05 mR=5cm=0.05m
Radius of each coil R = 5 cm =0.05 mR=5cm=0.05m Magnetic field produced by coil PP and QQ is B_1B1 and B_2B2 in xx and yy direction, respectively.
Radius of each coil R = 5 cm =0.05 mR=5cm=0.05m Magnetic field produced by coil PP and QQ is B_1B1 and B_2B2 in xx and yy direction, respectively.Magnetic field produced by coil P, B_1 = \dfrac{\mu_o I_1}{2R}B1=2RμoI1 Tesla \hat{x}x^
Radius of each coil R = 5 cm =0.05 mR=5cm=0.05m Magnetic field produced by coil PP and QQ is B_1B1 and B_2B2 in xx and yy direction, respectively.Magnetic field produced by coil P, B_1 = \dfrac{\mu_o I_1}{2R}B1=2RμoI1 Tesla \hat{x}x^ \therefore∴ B_1 = \dfrac{3 \mu_o }{2(0.05)} =30 \mu_oB1=2(0.05)3μo=30μo Tesla \hat{x}x^
Radius of each coil R = 5 cm =0.05 mR=5cm=0.05m Magnetic field produced by coil PP and QQ is B_1B1 and B_2B2 in xx and yy direction, respectively.Magnetic field produced by coil P, B_1 = \dfrac{\mu_o I_1}{2R}B1=2RμoI1 Tesla \hat{x}x^ \therefore∴ B_1 = \dfrac{3 \mu_o }{2(0.05)} =30 \mu_oB1=2(0.05)3μo=30μo Tesla \hat{x}x^Magnetic field produced by coil Q, B_2 = \dfrac{\mu_o I_2}{2R}B2=2RμoI2 Tesla \hat{y}y^
Radius of each coil R = 5 cm =0.05 mR=5cm=0.05m Magnetic field produced by coil PP and QQ is B_1B1 and B_2B2 in xx and yy direction, respectively.Magnetic field produced by coil P, B_1 = \dfrac{\mu_o I_1}{2R}B1=2RμoI1 Tesla \hat{x}x^ \therefore∴ B_1 = \dfrac{3 \mu_o }{2(0.05)} =30 \mu_oB1=2(0.05)3μo=30μo Tesla \hat{x}x^Magnetic field produced by coil Q, B_2 = \dfrac{\mu_o I_2}{2R}B2=2RμoI2 Tesla \hat{y}y^\therefore∴ B_2 = \dfrac{4 \mu_o }{2(0.05)} =40 \mu_oB2=2(0.05)4μo=40μo Tesla \hat{y}y^
Radius of each coil R = 5 cm =0.05 mR=5cm=0.05m Magnetic field produced by coil PP and QQ is B_1B1 and B_2B2 in xx and yy direction, respectively.Magnetic field produced by coil P, B_1 = \dfrac{\mu_o I_1}{2R}B1=2RμoI1 Tesla \hat{x}x^ \therefore∴ B_1 = \dfrac{3 \mu_o }{2(0.05)} =30 \mu_oB1=2(0.05)3μo=30μo Tesla \hat{x}x^Magnetic field produced by coil Q, B_2 = \dfrac{\mu_o I_2}{2R}B2=2RμoI2 Tesla \hat{y}y^\therefore∴ B_2 = \dfrac{4 \mu_o }{2(0.05)} =40 \mu_oB2=2(0.05)4μo=40μo Tesla \hat{y}y^Total magnetic field produced at the center B =\sqrt{B_1^2 + B_2^2}B=B12+B22
Answer:
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