Consider and are he currents flowing simultaneously in two nearby coils 1 and 2, respectively. If = self-inductance of coil 1, = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be [2025]

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Q.
Consider $I_{1}$ and $I_{2}$ are the currents flowing simultaneously in two nearby coils 1 and 2, respectively. If $L_{1}$ = self-inductance of coil 1, $M_{12}$ = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be [2025]

1 $ε_{1} = - L_{1} \frac{d I_{1}}{d t} + M_{12} \frac{d I_{2}}{d t}$

2 $ε_{1} = - L_{1} \frac{d I_{1}}{d t} - M_{12} \frac{d I_{1}}{d t}$

3 $ε_{1} = - L_{1} \frac{d I_{1}}{d t} - M_{12} \frac{d I_{2}}{d t}$

4 $ε_{1} = - L_{1} \frac{d I_{2}}{d t} - M_{12} \frac{d I_{1}}{d t}$

Ans.
(3)

$ϕ_{1} = L_{1} I_{1} + M_{12} I_{2}$

$ε_{1} = - \frac{d ϕ_{1}}{d t} = - L_{1} \frac{d I_{1}}{d t} - M_{12} \frac{d I_{2}}{d t}$

Magnitude of induced emf due to self-inductance, $= \frac{L d I_{1}}{d t}$

Magnitude of induced emf due to mutual inductance, $= \frac{M d I_{2}}{d t}$

Solution

Q.
Consider $I_{1}$ and $I_{2}$ are the currents flowing simultaneously in two nearby coils 1 and 2, respectively. If $L_{1}$ = self-inductance of coil 1, $M_{12}$ = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be [2025]

1 $ε_{1} = - L_{1} \frac{d I_{1}}{d t} + M_{12} \frac{d I_{2}}{d t}$

2 $ε_{1} = - L_{1} \frac{d I_{1}}{d t} - M_{12} \frac{d I_{1}}{d t}$

3 $ε_{1} = - L_{1} \frac{d I_{1}}{d t} - M_{12} \frac{d I_{2}}{d t}$

4 $ε_{1} = - L_{1} \frac{d I_{2}}{d t} - M_{12} \frac{d I_{1}}{d t}$

Ans.
(3)

$ϕ_{1} = L_{1} I_{1} + M_{12} I_{2}$

$ε_{1} = - \frac{d ϕ_{1}}{d t} = - L_{1} \frac{d I_{1}}{d t} - M_{12} \frac{d I_{2}}{d t}$

Magnitude of induced emf due to self-inductance, $= \frac{L d I_{1}}{d t}$

Magnitude of induced emf due to mutual inductance, $= \frac{M d I_{2}}{d t}$

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Solution

Q. Consider I1 and I2 are the currents flowing simultaneously in two nearby coils 1 and 2, respectively. If L1 = self-inductance of coil 1, M12 = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be [2025]

1 ε1=–L1dI1dt+M12dI2dt

2 ε1=–L1dI1dt–M12dI1dt

3 ε1=–L1dI1dt–M12dI2dt

4 ε1=–L1dI2dt–M12dI1dt

Ans. (3) ϕ1=L1I1+M12I2 ε1=–dϕ1dt=–L1dI1dt–M12dI2dt Magnitude of induced emf due to self-inductance, =LdI1dt Magnitude of induced emf due to mutual inductance, =MdI2dt

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Q.
Consider $I_{1}$ and $I_{2}$ are the currents flowing simultaneously in two nearby coils 1 and 2, respectively. If $L_{1}$ = self-inductance of coil 1, $M_{12}$ = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be [2025]

1 $ε_{1} = - L_{1} \frac{d I_{1}}{d t} + M_{12} \frac{d I_{2}}{d t}$

2 $ε_{1} = - L_{1} \frac{d I_{1}}{d t} - M_{12} \frac{d I_{1}}{d t}$

3 $ε_{1} = - L_{1} \frac{d I_{1}}{d t} - M_{12} \frac{d I_{2}}{d t}$

4 $ε_{1} = - L_{1} \frac{d I_{2}}{d t} - M_{12} \frac{d I_{1}}{d t}$

Ans.
(3)

$ϕ_{1} = L_{1} I_{1} + M_{12} I_{2}$

$ε_{1} = - \frac{d ϕ_{1}}{d t} = - L_{1} \frac{d I_{1}}{d t} - M_{12} \frac{d I_{2}}{d t}$

Magnitude of induced emf due to self-inductance, $= \frac{L d I_{1}}{d t}$

Magnitude of induced emf due to mutual inductance, $= \frac{M d I_{2}}{d t}$