Assertion (A): An emf is induced in a closed loop where magnetic flux is varied. The induced electric field is not a conservative field.
Reason (R): For induced electric field, the line integral \( \oint \vec{E} \cdot d\vec{l} \) around a closed path is non-zero.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Assertion (A) is true because varying magnetic flux induces a non-conservative electric field. Reason (R) is true; for a non-conservative field, \( \oint \vec{E} \cdot d\vec{l} \neq 0 \). (R) correctly explains why the induced electric field is non-conservative.
Assertion (A): The mutual inductance of two coils is doubled if the self-inductance of the primary and secondary coil is doubled.
Reason (R): Mutual inductance \(M \propto \sqrt{L_1 L_2}\).
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
The mutual inductance is given by \(M = k \sqrt{L_1 L_2}\). If \(L_1\) and \(L_2\) are doubled, the new mutual inductance becomes \(M' = k \sqrt{(2L_1)(2L_2)} = 2k \sqrt{L_1 L_2} = 2M\). Thus, (A) is true and (R) correctly explains (A).
Assertion (A): A system cannot have mutual inductance without having self inductance.
Reason (R): If mutual inductance of system is zero, its self-inductance must be zero.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Assertion (A) is true, as mutual inductance \(M\) depends on the self-inductances \(L_1\) and \(L_2\) (\(M \le \sqrt{L_1 L_2}\)). Reason (R) is false; if two coils are perfectly uncoupled (\(k=0\)), their mutual inductance is zero, but their self-inductances \(L_1\) and \(L_2\) can still be non-zero.