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Show that a bar magnet behaves as an equivalent current carrying solenoid. |
Answer» Solution : Let .a. be radius of solenoid of length 2l. To calculate magnetic field at a point on the a x is of a solenoid, consider a small element of thickness ‘dx. of solenoid at a distance .x. from .O.. Number of TURNS in this element =n.dx If current .l. flows through element n dx the magnitude of magnetic field at ‘P. due to this element is `dB = (mu_0)/(4pi) (2pi (ndx) Ia^2)/([(r -x)^2 + a^2]^(1//2) )` If point .P. is at large distance from .O. i.e., r >>x and r >> a then `[(r -x)^2 +a^2] = r^2` `dB = (mu_0)/(4pi) (2pi (ndx) Ia^2)/((r^2)^(3//2))` ` dB = (mu_0)/(4pi) (2pi (ndx) Ia^2)/(r^3)` The TOTAL magnetic field at .P. due to the current .I. in solenoid is `B = int_(-1)^1 dB = int_(-1)^(1) (mu_0)/(4pi) (2pi n I a^2 dx)/(r^3)` `B = (mu_0)/(4pi) (2pi nIa^2)/(r^3) int_(-1)^1 dx ` `B = (mu_0)/(4pi) (2pi nIa^2)/(r^3) [x]_(-1)^1` `B = (mu_0)/(4pi) (2pi nIa^2)/(r^3) [l+l]` `B = (mu_0)/(4pi) (2In pi a^2 2l)/((r )^3)` `B = (mu_0)/(4pi) (2In A.2l)/(r^3) "" ( because pia^2 = A)` `B = (mu_0)/(4pi) (2(n.2l)IA)/(r^3)` `B = (mu_0)/(4pi)( 2NIA)/(r^3)` `B = (mu_0)/(4pi) (2m)/(r^3)` This equation is similar to expression for magnetic field on the AXIS of a short BAR magnet. HENCE a solenoid carrying current behaves as a bar magnet. |
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