A flux of \(10^{-3}\text{ Wb}\) passes through a strip having an area \(A = 0.02\text{ m}^2\). The plane of the strip is at an angle of \(60^\circ\) to the direction of a uniform field \(B\). The value of \(B\) is:
1. \(0.1\text{ T}\)
2. \(0.058\text{ T}\)
3. \(4.0\text{ mT}\)
4. None of the above
View Answer
The magnetic flux through the surface is given by \(\phi = B A \sin\theta\). Substituting \(\phi = 10^{-3}\text{ Wb}\), \(A = 0.02\text{ m}^2\), and \(\theta = 60^\circ\), we get \(B = \frac{10^{-3}}{0.02 \times \sin 60^\circ} \approx 0.058\text{ T}\).
Assertion (A): The electric field created by time-varying magnetic field is non-conservative.
Reason (R): The line integral of induced electric field in a closed loop is always equal to zero.
1. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
2. Both Assertion and Reason are true but Reason is not correct explanation of Assertion.
3. Assertion is true but Reason is false.
4. Assertion and Reason are false.
View Answer
Electric field created by time-varying magnetic field is non-conservative, which means its line integral around a closed loop is non-zero (\(\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}\)). Hence, Assertion is true but Reason is false.
Two conducting circular loops of radii \(R_1\) and \(R_2\) are placed in the same plane with their centres coinciding. If \(R_1 \gg R_2\), the mutual inductance \(M\) between them will be directly proportional to
1. \(\frac{R_2^2}{R_1}\)
2. \(\frac{R_1}{R_2}\)
3. \(\frac{R_2}{R_1}\)
4. \(\frac{R_1^2}{R_2}\)
View Answer
The magnetic field produced by the larger loop 1 at the center is \(B_1 = \frac{\mu_0 I_1}{2 R_1}\). The magnetic flux through the smaller loop 2 is \(\phi_2 = B_1 A_2 = \frac{\mu_0 I_1}{2 R_1} \pi R_2^2\). Therefore, \(M = \frac{\phi_2}{I_1} = \frac{\mu_0 \pi R_2^2}{2 R_1}\), which means \(M \propto \frac{R_2^2}{R_1}\).
A step down transformer connected to an ac mains supply of \(220\text{ V}\) is made to operate at \(11\text{ V}\), \(44\text{ W}\) lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?
1. 4 A
2. 0.2 A
3. 0.4 A
4. 2 A
View Answer
Assuming zero power losses, the input power in the primary circuit equals the output power in the secondary circuit: \(P_{\text{in}} = P_{\text{out}} = 44\text{ W}\). Since \(P_{\text{in}} = V_p I_p\), we get \(I_p = \frac{P_{\text{in}}}{V_p} = \frac{44}{220} = 0.2\text{ A}\).