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Along wire carries a current of 18 A kept along the axis of a long solenoid of radius 1 cm. The field due to the solenoid is 8.0 xx10^(-3)T. The magnitude of the resultant field at a point 0.6 mm from the solenoid axis is (Assume mu_0 = 4pi x 10^(-7)Tm/A) |
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Answer» `6 xx 10^(-3)T`  Magnetic field due to a solenoid , ` B_1 = 8 xx 10^(-3) T` and r = 0.6 mm ` = 0.6 xx 10^(-3) m` Radius of solenoid , `R = 1 cm = 1 xx 10^(-2) m ` Magnetic field due to current carrying wire at a point , P `B_2 = (mu_0)/(2pi) . I/r = 2 xx 10^(-7) xx ( 18)/(0.6 xx 10^(-3) )` ` = 6 xx 10^(-3) T `(in upward direction) Since, current carrying wire is placed along the axis of solenoid, HENCE magnetic field produced by wire and magnetic field due to solenoid at point P, both are perpendicular to each other, Hence, the magnitude of the resultant magnetic field, `B = sqrt (B_1^2 + B_2^2)` ` = sqrt ( (8 xx 10^(-3))^2 + (6 xx 10^(-3) )^2 )` ` = 10 xx 10^(-3) T` |
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