Q4. Assertion: Ampere’s circuital law cannot be used to find the field at the centre of a circular current loop. Reason: the law is true only for loops of high symmetry. Are both true, and does the reason explain the assertion?
The assertion is true but the reason is false. The law holds for every closed loop; at the centre of a circular loop there is simply no amperian loop on which B is tangential and constant, so the law does not give B there easily (NCERT page 120).
Q5. Two long parallel wires 10 cm apart carry 4.0 A and 6.0 A in the same direction. Find the field midway between them.
Each gives B = μI/(2πr) at r = 0.050 m: 1.6 × 10−5 T and 2.4 × 10−5 T. With currents in the same direction the two fields at the midpoint point opposite ways, so the net field is 8.0 × 10−6 T, in the direction of the field of the 6.0 A wire (NCERT Eq. (4.14) and the right-hand rule, page 119).
Q6. A long hollow metal pipe carries a current along its length, spread evenly round its wall. What is the field inside the pipe, and outside it?
In the hollow cavity (not in the conducting wall), with no other sources present, an amperian circle encloses no current and cylindrical symmetry makes B tangential and constant, so B = 0. Outside, at distance r from the axis, the whole current is enclosed and B = μI/(2πr), as for a thin wire (reasoned from NCERT Eq. [4.13(b)], page 118).
Q7. In a thick wire of radius a carrying a uniform current, at what distance from the axis inside the wire is the field half its value at the surface?
Inside, B ∝ r, so B = B(a)/2 at r = a/2. Outside it falls as 1/r and is half the surface value again at r = 2a (NCERT Example 4.7, page 120).
Q8. In a long wire of radius a, the current density rises in proportion to the distance from the axis, J = kr, and the total current is I. Find B inside the wire.
A thin ring of radius x and width dx carries kx × 2πx dx. Adding the rings from 0 to r gives 2πkr3/3, so the enclosed current grows as r3: Ie = Ir3/a3. Then B × 2πr = μ0Ir3/a3, so B = μ0Ir2/(2πa3). It rises as r2 and meets μ0I/(2πa) at the surface (beyond NCERT; reasoned from NCERT Eq. [4.13(b)], page 118).