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

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,

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

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

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

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

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

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