Question

The magnetic field at the centre of given coil having radius 2cm is (current is 100amp)

Answer 1

Answer:

At the centre of the coil, x=0x=0x=0.

Hence, magnetic field at centre is Bc=μoIR2N2R3B_c=\cfrac{\mu_oIR^2N}{2R^3}Bc​=2R3μo​IR2N​

B=μoNI2RB=\cfrac{\mu_o NI}{2R} B=2Rμo​NI​

(b)

In a small region of length 2d about the mid-point between the coils,

B = μ0 IR2N2 × [{(R2 + d)2 + R2}−3/2 + {(R2 − d )2 + R2}−3/2] B\, =\, \displaystyle \frac {\mu_0\,IR^2N}{2}\, \times\, \left [ \left \{ \left ( \displaystyle \frac{R}{2}\, +\, d \right )^2\, +\,R^2 \right \}^{-3/2}\, +\,\left \{ \left (\displaystyle \frac {R}{2}\,-\, d  \right )^2\, +\, R^2 \right \}^{-3/2} \right ] B=2μ0​IR2N​×⎣⎢⎡​{(2R​+d)2+R2}−3/2+{(2R​−d )2+R2}−3/2⎦⎥⎤​

= μ0 IR2N2 × (5R24)−3/2 × [(1 + 4d5R) + (1 − 4d5R)−3/2]=\, \displaystyle \frac {\mu_0\,IR^2N}{2}\, \times\,\left ( \displaystyle \frac {5R^2}{4} \right )^{-3/2}\, \times\, \left [ \left ( 1\, +\, \displaystyle \frac {4d}{5R} \right )\, +\, \left ( 1\, -\, \displaystyle \frac {4d}{5R} \right )^{-3/2} \right ]=2μ0​IR2N​×(45R2​)−3/2×[(1+5R4d​)+(1−5R4d​)−3/2]

= μ0 IR2N2 × (45)−3/2 × [1 − 6d5R + 1 + 6d5R]=\, \displaystyle \frac {\mu_0\,IR^2N}{2}\, \times\,\left ( \displaystyle \frac {4}{5} \right )^{-3/2}\, \times\,\left [1\, -\, \displaystyle \frac {6d}{5R}\,+\,1\,+\, \displaystyle \frac {6d}{5R} \right ]=2μ0​IR2N​×(54​)−3/2×[1−5R6d​+1+5R6d​]

where in the second and third steps above, terms containing d2/R2d^2/R^2d2/R2 and higher powers of d/R are neglected since dR << 1\displaystyle \frac {d}{R}\, <<\, 1Rd​<<1 The terms linear in d/R cancel giving a uniform field B in a small region:

B = μ0 IR2N2 × (45)−3/2 μ0INR = 0.72 μ0INRB\,=\, \displaystyle \frac {\mu_0\,IR^2N}{2}\, \times\,\left ( \displaystyle \frac {4}{5} \right )^{-3/2}\,\displaystyle \frac {\mu_0IN}{R}\, =\, 0.72\, \displaystyle \frac {\mu_0IN}{R}B=2μ0​IR2N​×(54​)−3/2Rμ0​IN​=0.72Rμ0​IN​

Explanation: