Question

A uniform circular wire loop is connected to the terminals of a battery. The magnetic field induction at the centre due to ABC portion of the wire will be (length of ABC =L1, length of ADC= L2)

Answer 1

"The magnetic field at the center of the circular loop that is uniform in nature = 0

Here, it is to find on part of the loop so let the current that is flowing through ABC be i1 and ADC be i2.

Both are parallel in nature,

i1/i2 = R2/R1

i1/i2 = ρl2/A divided by ρl1/2

i1/i2 = l2/l1

i1 = [l2/ (l1 + l2)] i2 will be eq (1)

in ABC magnetic field,

ui i1 thetha / 4*pie*r eq (2)

loop at centre,

thetha/2pie – thetha = liR/l2R

thetha = l1 /(l1+l2) * 2pie eq (3)

put (1) and (2) in (3)

magnetic = ui I l2 * l1 2pie/ 4pie*r (l1+l2) (l1+l2)

B = ui I l1 l2 / 2R (l1+L2)^2

"

Answer 2

Answer:

Magnetic field induction at the centre due to ABC portion of the wire will be =>

$B = \frac{\mu_ol_1l_2i}{2R {(l_1 +l_2)}^{2} }$

Explanation:

According to the question;

Length of ABC = $l_1$

Length of ADC = $l_2$

Let current in part ABC be $i_1$

& in part ADC be $i_2$

As ABC and ADC part are in parallel connection;

Therefore,

$= > \frac{i_1}{i_2} = \frac{R_1}{R_2} \\ \\ (R_1 = \frac{\rho \: l_1}{A} \: \: ; R_2 = \frac{\rho \: l_2}{A}) \\ \\ = > \frac{i_1}{i_2} = \frac{\frac{\rho \: l_2}{A}}{\frac{\rho \: l_1}{A}} \\ \\ = > \frac{i_1}{i_2} = \frac{l_2}{l_1} \\ \\ = > i_2 = \frac{l_1i_1}{l_2}$

$Sum \: of \: i_1 \: and \: i_2 \: is \: equal \: to \\ total \: current \: in \: the \: circuit \: i.e. \: i \\ \\ = > i_1 + i_2 = i \\ \\ = > i_1 + \frac{l_1i_1}{l_2} = i \\ \\ = > ( \frac{l_2 +l_1 }{l_2} )i_1 = i \\ \\ = > i_1 = (\frac{l_2}{l_2 +l_1} )i$

$Radius: \\ \\ 2\pi \: R = l_1 +l_2 \\ \\ = >R = \frac{(l_1 +l_2 )}{2\pi}$

$Angle \: subtended \: by \: ABC \: at \\ center \: O \: will \: be (\theta) \: = \frac{Arc}{Radius} \\ \\ (Arc = l_1)$

$= > \frac{l_1}{\frac{(l_1 +l_2 )}{2\pi}} \\ \\ = > \frac{2\pi \: l_1}{l_1 +l_2}$

$Magnetic \: field \: due \: to \: a \\ section \: of \: ring \: (i.e. \: length \\ ABC); \\ \\ B = \frac{\mu_oi_1}{2R} ( \frac{\theta}{2\pi} ) \\ \\ B = \frac{\mu_o}{2R} (\frac{l_2i}{l_2 +l_1} ) (\frac{2\pi \: l_1}{l_1 +l_2})\times \frac{1}{2\pi} \\ \\ B = \frac{\mu_ol_1l_2i}{2R {(l_1 +l_2)}^{2} }$