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Derive equation of magnetic field inside a long straight solenoid. |
Answer» Solution :1. Figure shows the sectional view of a long solenoid. At various turns of the solenoid, current COMES out of the plane of paper at points marked  and enters the plane of paper at points marked  .  2. To determine the magnetic field `vecB` at any inside point, consider a rectangular closed path abcd as the Amperean LOOP. 3. According to Ampere.s circuital law, `ointvecB*vec(dl)=int_(a)^(b)vecB*vec(dl)+int_(b)^(c)vecB*vec(dl)+int_(c)^(d)vecB*vec(dl)+int_(d)^(a)vecB*vec(dl)""...(1)` 4. cd part is outside the solenoid. Outside magnetic field `|vecB|=O` so that, `int_(c)^(d)vecB*vec(dl)=O` 5. d -a and b - c parts are perpendicular to `vecB` so, `int_(b)^(c)vecB*vec(dl)=int_(d)^(a)vecB*vec(dl)=O` 6. So equation (1) will be like this, `ointvecB*vec(dl)=int_(a)^(b)vecB*vec(dl)=int_(a)^(b)BdlcosO=Bint_(a)^(b)dl` `thereforeointvecB*vec(dl)=B(h)""...(2)`(Where h = length of part ab) 7. SUPPOSE, number of TURN per unit length be n. So total number nh in h length. 8. I be current to each turn, so nhI current for nh turns current enclosed by loop `I_(e)=I(nh)""...(3)` 9. According to Ampere.s circuital law, `ointvecB*vec(dl)=mu_(0)I_(e)""...(4)` `therefore` Substitute equation (2) and (3) in equation (4), `B(h)=mu_(0)Inh` `thereforeB=mu_(0)nI""...(5)` 10. The direction of the field is given by the right hand rule. 11. The solenoid is commonly used to obtain uniform magnetic field. |
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